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EMERSON’S THIRD PART. 


THE 


NORTH AMERICAN ARITHMETIC. 


‘PART THIRD, 
FOR 


ADVANCED SCHOLARS. 


BY FREDERICK EMERSON, 


LATE PRINCIPAL IN THE DEPARTMENT OF ARITHMETIC, 
BOYLSTON SCHOOL, BOSTON. 


| BOSTON: 
RUSSELL, ODIORNE, & METCALF. 


NEW YORK, Collins & Hannay. PHILADELPHIA, Hogan & Thomp- 


son. BALTIMORE, David Cushing. H ARTFORD, Pdi Huntington. 
WINDSOR, Ide & Goddard. HALLOWELL, Glazier, Masters, & Co. 
CINCINNATI, C, P. Barnes. RALEIGH, Turner & Hughes. 


183.45" & 


Entered, according to act of Congress, in the year 1834, by FrepErick 
KareRsox, in the Clerk’s office of the District Court of the District of 
Massachusetts. 


STEREOTYPED BY THOMAS G. WELLS AND CO. 
BOSTON, 


a — Say 


eMATIOS Llbieyhs 


PREFACE. 


Tue work now presented, is the last of a series of books, 
under the general title of ‘Tue Norta American ArirH- 
mETIC, and severally denominated Part First, Part Second, 
and Part Third. 3 

Part First isa small book, designed for the use of child- 
ren between five and eight years of age, and suited to the 


_ convenience of class-teaching in primary schools. 


| 


Parr Seconp consists of a course of oral and written ex- 
ercises united, embracing sufficient theory and practice of 
arithmetic for all the purposes of common business. 

Part Tuirp comprises a brief view of the elementary 
principles of arithmetic, and a full development of its higher 
operations. Although it is especially prepared to succeed 
the use of Part Second, it may be conveniently taken up by 
scholars, whose acquirements in arithmetic are considerably 


- less than the exercises in Part Second are calculated to ai- 


ford. While preparing this book, I have kept in prominent 


» view, two classes of scholars; viz.— those who are to prose- 


a 


cute a full course of mathematical studies, and those who 
are to embark in commerce. In attempting to place arith- 


_ metic, as a science, before the scholar in that light, which 


shall prepare him for the proper requirements of college, 1 
have found it convenient to draw a large portion of the ex- 
amples for illustration and practice, from mercantile trans- 
actions; and thus pure and mercantile arithmetic are united. 
No attention has been spared, to render the mercantile 
information here presented, correct and adequate. Being 


_ convinced, that many of the statements relative to commerce, 


which appear in books of arithmetic, have been transmitted 
down from ancient publications, and are now erroneous, I 
have drawn new data from the counting-room, the insurance 
office, the custom-house, and the laws of the present times. 
The article on Foreign Exchange is comparatively exten- 
sive, and I hope it will be found to justify the confidence of 
merchants, Its statements correspond to those ofthe British 
‘Unwersal Cambist,’ conformably with our value of foreign. 
coins, as fixed by Act of Congress, in 1834. 


BO3134 


4 PREFACE. 


Although a knowledge of arithmetic may, in general, be | 


well appreciated as a valuable acquisition, yet the effect 
produced on intellectual character, by the exercises neces- 
sary for acquiring that knowledge, is not always duly con- 
sidered. In these exercises, the mental effort required in 
discovering the true relations of the data, tends to strengthen 
the power of comprehension, arfd leads to a habit of investi- 
gating; the certainty of the processes, and the indisputable 
correctness of the results, give clearness and activity of 
thought; and, in the systematic arrangement necessary to 
be observed in performing solutions, the mind is disciplined 
to order, and accustomed to that connected view of things, 
so indispensable to the formation of a sound judgment. 
These advantages, however, depend on the ‘manner in 
which the science is taught; and they are gained, or lost, 
in proportion as the teaching is rational, or superficial. 

Arithmetic, more than any other branch of learning, has 
- suffered from the influence of circumstances. Being the 
vade-mecum of the shop-keeper, it has too often been 
viewed as the peculiar accomplishment of the accountant, 
and neglected by the classical student. The popular sup- 
position, that a compendious treatise can be more easily 
mastered than a copious one, has led to the use of text- 
books, which are deficient, both in elucidation and exer- 
cises. But these evils seem now to be dissipating. —The 
elements of arithmetic have become a subject of primary 
instruction; and teachers of higher schools, who have adopt- 
ed an elevated course of study, are no longer satisfied with 
books of indifferent character. 

It has been my belief, that a treatise on arithmetic might 
_be sé constructed, that the learner should find no means of 
proceeding in the exercises, without mastering the subject 


in his own mind, as he advances; and, that he should still be . 


enabled to proceed through the entire course, without requir- 
ing any instruction from his tutor. Induced by this belief, 
I commenced preparing The North American Arithmetic 
about five years since; and the only apology I shall offer, 
for not earlier presenting its several Parts to the public, is 
the unwillingness that they should pass from my hands, 
while I could see opportunity for their improvement. 


Boston, October, 1834. FE’. EMERSON. 


A KEY to this work (for teachers only) is published separately. 


ws 


1. 


ARITHMETIC. 


ARTICLE I. 


DEFINITIONS OF QUANTITY, NUMBERS, AND 
ARITHMETIC. 


QUANTITY is that property of any thing which may 
be increased or diminished—it is magnitude or multi- 
tude. It is magnitude when presented in a mass or con- 
tinuity ; as, a quantity of water, a quantity of cloth. It 
is multitude when presented in the assemblage of several 
things ; as, a quantity of pens, a quantity of hats. The 
idea of quantity is not, however, confined to visible ob- 
jects ; it has reference to avery thing that is susceptible 
of being more or less. 

NUMBERS are the expressions of quantity. Their 

names are, One, Two, Three, Four, Five, Six, Seven, 
Eight, Nine, Ten, &c. In quantities of multitude, One 
expresses a Unit; that is, an entire, single thing; as 
one pen, one hat. Then each succeeding number ex- 
presses one unit more than the next preceding. In 

quantities of magnitude, a certain known quantity is first 
assumed as a measure, and considered the unit; as one 
gallon, one yard. Then each succeeding number ex- 
presses a quantity equal to as many times the unit, as the 
number indicates. » Hence, the value of any number de- 
pends upon the nice of-its unity. 

When the unit is applied to any particular thing, it is 
called a concrete unit ; and pe consisting of concrete 


« 


6 ARITHMETIC. | II. 


units are called concrete numbers: for example, one dollar, 
two dollars. But when no particular thing is indicated 
by the unit, it is an abstract unit; and hence arise abstract 
numbers: for example, one and one make two. jens 

Without the use of numbers, we cannot know precise- 
ly how much- any quantity is, nor make any exact com- 
parison of quantities. And it is by comparison only, that 
we value all quantities; since an object, viewed by itself, 
camot be considered either great or small, much or hit- 
tle; it can be so only in its relation to some other object, 
that is smaller or greater. | 

ARITHMETIC treats of numbers: it demonstrates 
their various properties and relations; and hence it 1s 
called the Science of numbers. It also teaches the 
methods of computing by numbers; and hence it is call- 
ed the Art of numbering. 


iH, 
NOTATION AND NUMERATION. 


Notation is the writing of numbers im numerical char- 
acters, and NumERATION is the reading of them. 

The method of denoting numbers first practised, was 
undoubtedly that of representing each unit by a separate 
mark. Various abbreviations of this method succeeded; 
such as the use of a single character to represent five, 
another to represent ten, &c.; but no method was found 
perfectly convenient, until the Arabic FIGURES or DIGITS, 
and DECIMAL system now in use, were adopted. These 


figures are, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; denoting respec- — 


- 


tively, nothing, one unit, two units, three units, &c. 
To denote numbers higher than 9, recourse is had to a 
law that assigns superior values to figures, according to the 
order in which they are placed. viz. Any figure placed 
to the left of another figure, expresses ten times the quantity 
that it would express if it occupied the place of the latter. 
Hence arise a succession of higher orders of units. 
As an illustration of the above law, observe the dif- 


ferent quantities which are expressed by the figure 1. — 
: > 


watt 


- 


‘ 


I. NOTATION AND NUMERATION. 7 


» When standing alone, or to the right of other figures, 1 
represents 1 unit of the first degree or order; when stand- 
ing in the second place towards the left, thus, 10, it 
represents 1 ten, which is 1 unit of the second degree ; 
when standing in the third place, thus, 100, it represents 
1 hundred, which is 1 unit of the third degree: and so on. 
The zero or cipher (0) expresses nothing of itself, yore 
employed only to occupy a place. jb ay 
The units of the second degree, that is, the tens, are 
denoted and named in succession, 10 ten, 20 twenty, 30 
thirty, 40 forty, 50 fifty, 60 sixty, 70 seventy, 80 eighty, 
90 ninety. The units of the third degree, that is, the 
hundreds, are denoted and named, 100 one hundred, 200 
_ two hundred, 300 three hundred, and so on to 900 nine 
hundred. The numbers Benen 10 and 20 are denoted 
and named, 11 eleven, 12 twelve, 13 thirteen, 14 four- 
teen, 15 fifteen, 16 sixteen, 17 seventeen, 18 eighteen, 
/19 nineteen. Numbers between all other tens are de- 
noted in like manner, but their names are compounded of 
the names of their respective units; thus, 21 twenty-one, 
22 twenty-two, 23 twenty-three, &e.: ; ol thirty-one, 32 
thirty-two, &c. &c. This nomenclature, although not 
very imperfect, might be rendered more consistent, by 
substituting regular compound names for those now ap- 
plied to the numbers between 10 and 20. This alter- 
ation would give the names, 11 ten-one, 12 ten-two, 13 
ten-three, &e. 

As the first three places of figures are appropriated to 
simple units, tens, and hundreds, so every succeeding three 
places are appropriated to the units, tens, and hundreds 
of succeeding higher denominations._ For illustration, see 
the following table. 


a 

S a a wn p 

2 Ss = u re 2 oa ua 
=e ae ae ee ae 

oO S) ° ° oS 2 aig ae = Fal n a se 

© iS) as, = et = oe & ra q 5 we 

ee tr MEE ee et Caco « 
BRE Ae pe = ae 
~ @ RES Siete: 
oS 70 ite Can # @ eH ® BR SB 
aa (FRING Gea Ek ON Coe DR GO EK ee ON 
450725 206 194 007 185 039 000 164 396 205 013008741 


By continuing to adopt a new name for every three 
degrees of units, the above table may be extended indef- 


- . 


8g _ > ARITHMETIC. IL 


initely. Formerly, the denominations higher than thousand; 
were eachmade to embrace six degrees of units; taking i inf 
thousands, tens of thousands, and hundreds of thousands b 
The mode of applying a name to every three degrees’ 
however, is now universal on the continent of Europe b 


and is becoming so in England and America. 


e 


The learner may denote in figures, the following num} 
bers, which are written in words. | 

Example 1. Four hundred seventy-eight million, tw¢ 
hundred forty-one thousand, and one hundred. 

2. Seven million, six hundred ninety-two thousand _ 
and eighty-nine. 

3. Nineteen million, twenty thousand, and five. 

4. Eight hundred billion. | 

5. One billion, six hundred forty-four thousand, five 
hundred and thirteen. 

6. One trillion, five hundred thirty-four billion, thred 
million, eighteen thousand, and four. _ ; 
7. Two hundred Pian: sixteen thousand and one. 

8. Eleven billion, one million, and sixty. 
9. Five trillion, eight billion, four million, nine thou 
sand, and seven. | 
10. One hundred trillion, twenty billion, thr ee hun) 
dred million, two thousand, and four. 7 ae 
11. Thirty-one trillion, five hundred, and sixty. : 
12. Six quadrillion, two hundred ‘and ‘fourteen trillion! 
13. Two hundred forty-nine quadrillion, seventy-fivi 
thousand, and twenty-two. 
14. Forty-six quintillion, one quadrillion, nmeteen bil 
ion, seven hundred and eight. | 
15. Nine hundred sextillion, three hundred twenty, 
five trillion, two thousand, and fourteen. ) | 


INDICATIVE CHARACTERS OR SIGNS. 


The sign-+ (plus) between numbers, indicates a 
they are to be added together; thus, 3-++ 2 is 5. 

The sign— (minus) mdicates, that the number placer 
after it, is to be subtracted from the number placed be 
fore it; thus, 5—2 is 3. 


Lf. ADDITION. S 


_ The sign X (into) indicates that one number is to be 
multiplied into another; thus, 4 x 2 is 12. 

_ Whe sign + (by) indicates that the number on the left 
hand is to be divided by the number on the right hand; 
ee 12-3 is 4oc 

_ The sign== (equal to) indicates that the number before 
it, is equal to the number after it; for example, 4+ 2— 6. 
@ 2-4. 5X3—15. 15+3=5. 


iil. 
ADDITION. 


ADDITION is the operation by which two or more num- 
bers are united in one number, called their sum. It is the 
first and most simple operation in arithmetic, effecting the 
first and most simple combination of quantities. 

The primary mode of forming numbers, by joining one 
unit to another, and, this sum to another, and. so on, ex- 
hibits the principle of addition. When numbers, which 
are to be added, consist of units of several degrees, such 
as tens, hundreds, &c., it is found. convenient to add 
together the units of each degree by themselves; and 
since ten units of any degree make one unit of the next 
higher degree, the number of tens in the sum of each 
degree of units is carried to the next higher degree, and 
added thereto. , é' 

RULE FOR ADDITION. Write the numbers, units un- 
der wnits, tens under tens, &c. Add each column sep- 
wately, beginning with-the column of wnits. When the 
sum of any column is not more than 9, write it under 
‘he column: when the sum is more than 9, write only the 
units’ figure under the column, and carry the tens to the 
wext column. Finally, write down the whole sum of the 
jejt hand column. es 


| 1. What is the sum of 370+ 90264-+ 1470+ 40060? 
_ 2. What is the sum of 4000-++ 570-+99-+ 54-273 + 
59073-+-4000-+-61998 +752? 
| 


10 ARITHMETIC. Iv 


3. What is the sum of 243-+- 5021 +-7628-+- 927+ 6. 
+5823-+742-+796 + 5009 -+ 325 +7426 +31186 4 
987-+-6954 +2748? | 

4. What is the sum of two thousand and seven, forty 
four million five hundred and sixty-one, one hundred mu) 
lion, six billion twenty-eight thousand and eleven ? 


YZ 
SUBTRACTION. | 


SuBTRACTION is the operation by which one numbe 
is taken from another. 

The number from which another is to be taken | 
called the minuend, and the number to be taken is calle 
the subtrahend. The number resulting from the ope: 
ation shows the remainder of the minuend, after th 
subtrahend has been taken out; it also shows the diffe 
ence between the minuend and subtrahend, or the exce, 
of the former above the latter. The subtrahend and ré 
mainder may be considered the two parts into which tk 
minuend is separated by the operation; and in this viev 
subtraction is the opposite of addition, in as much 
addition unites several quantities in one sum, and subtra 
tion separates a quantity into two parts. | 

Subtraction is performed by taking the units of eac 
degree in the subtrahend, from those of correspondir| 
degree in the minuend, and severally denoting the ri 
mainders. When the units of any degree in the subir 
hend exceed those of the same’ degree in the minuen 
we mentally join one unit of the next higher degree 
the deficient place in the subtrahend, and consider tl 
units of the higher degree to be one less than they a’ 
dencted: this process is the reverse of carrying , 
addition. One other method may be adopted in th, 
case; viz. Increase both the minuend and subtrahen 
by mentally adding ten to the deficient place in t) 
former, and, one to the next higher degree of units | 
the latter. This method is justified by the self-evide) 


|W: MULTIPLICATION. il 


truth, that, if two unequal quantities be equally increased, 
> their difference is not thereby altered. 

RULE FOR SUBTRACTION. Write the smaller number 
| under the greater, placing units under wnits, §c. Be- 
| gin with the units, and subtract each figure in the lower 
number from the figure over it. When a figure in the 
upper number is smaller than the figure under it, consid- 
er the upper figure tv be 10 more than it is, and the neat 
upper figure on the left hand, to be 1 less than it is. 

proor. Add together the remainder and the smaller 
number: their sum will be equal to the greater number, 
if the work be right. 


_ 4. What is the difference between 70240 and 69418? 
' 2. How much is the excess of the number 482724 
’ above the number 194750? 

' 3. Suppose 479021 to be a minuend, and’ 38456 
' the subtrahend; how much is the remainder? 

' 4. 905106392 — 904623724 —>? 

‘5. Subtract fifty-one thousand from one hundred bil- 


* lion, cighteen thousand, five hundred and one. 
i 


j et ening FE eh oe EE a SES ONT cee Le 


| MULTIPLICATION. 


}) 
¢ t 


MULTIPLICATION is the operation by which a number 
is produced, equal to as many times one given number, 
as there are units in another given number. It is an 
| abridged method of finding the sum of several equal 
| quantities, by repeating one of those quantities. 

| The number to be multiplied or repeated is called the 
| multiplicand; it may be viewed as one of several equal 
| quantities, whose sum is to be produced by the operation. 
{The number to multiply by 1s called the multiplier; 
it indicates how many such quantities as the multiplicand 
' are to be united, or, how many times the multiplicand is 
' to be repeated. Thenumber resulting from the operation 
| 3s called the product. 


12 ARITHMETIC. Vv 


The multiplicand and multiplier, considered as con: 
curring to form the product, are called factors of the 
product. Either factor may be used as the multiplier 
of the other; that is, the multiplicand and multiplier may 
change places, and the product will be still the same, 
For example, 4x 3=12. 3xX4=—12. 

When a product arises from more than two factors, . 
the numbers may be denoted thus, 6 x 3 x 5=90; but. 
in forming the product, a distinct operation is necessary te 
bring in each factor, after the two first. The numbers, 
6, 3, 5, would, therefore, be multiplied into each other 
tins, 6X 3—=I18: 18 x 5=90. 

Factors may be arranged in any succession whatever, 
since the mere order in which they are brought imto the, 
operation cannot affect their final product. For exam- 
ple, 5X3 xX4=60. 4X38xX5=60. 3X5X4=60, 

The products of small numbers may be committed to) 
memory; but when the product of factors consisting of 
several figures is required, it is necessary to multiply’ 
each figure in the multiplicand by each figure in the 
multiplier, and denote the several products in such ordei, 
that they shall represent their respective values. When 
simple units are employed as the multiplier, the product 
_ of each figure in the multiplicand is of the same degree! 
as the figure multiplied; that is, units multiplying units 
give units, units multiplying tens give tens, units multi- 
plying hundreds give hundreds, &c. When tens are 
employed as the multiplier, the product of each figure 
in the multiplicand is one degree higher than the figure 
multiplied; that is, tens multiplying units give tens, tens| 
multiplying tens give hundreds, tens multiplying hundreds 
give thousands, &c. When hundreds are employed as 
the multiplier, the product of each figure in the multi- 
plicand is two degrees higher than the figure multiplied; 
and so on. 

RULE FOR MULTIPLICATION. Write the multiplier 
under the multiplicand, placing units under units, §e. | 

When there is but one figure in the multiplier, begin 
with the units, multiply each figure in the multiplicand 
separately, and place each product under the figure im 


co 


| ¥, MULTIPLICATION. 13 


the multiplicand from which it arose; observing to carry 
_ the tens to the left as in addition. 

— - When there is more than one figure in the multiplier, 
multiply by each figure separately, and write its product 
ina separate line, placing the right hand figure of each 
line under the figure by which you multiply; and finally, 
add together the several products. The sum will be ° 
the whole product. 

__ Abbreviations of the above rule may frequently be 
adopted, as follows. 

When there are ciphers standing between other fig- 
ures, in the multiplier, they may be disregarded. 

When ciphers stand on the right of either factor, or 
both, they may be disregarded till the multiplication is 
performed, and then annexed to the product. 

When either factor is 10, 100, 1000, &c., merely 
place the ciphers in this factor on the right hand of the 
other factor, and it becomes the product. 

When the multiplier is a number that can be produc- 
ed by multiplying two smaller numbers together, multi- 
ply the multiplicand first by one of the smaller numbers, 
and the product thence arising by the other. 


1. Suppose 479265 to be a multiplicand, and 9236 
the multiplier; how much is the product ? 

2. Suppose 26537 to be one factor, and 873643 
another; how much is their product ? 

3. Suppose the numbers 725, 38046 and 91, to be 
factors; how much is the product? 

4. What is the product of 62392 x 4003 ° 

5. What is the product of 248000 < 9400? 

6. What is the product of 24 x 300 X13 10002? 

7. Multiply one hundred five million, by one thousand. 


For the purpose of determining whether any error has 
happened in the process of multiplication, the following 
' method of trial, which depends on the peculiar property 
of the number 9, and which is called casting owt the 
nines, may be practised. 

Add together the figures of the product, herizontally, 


14 ARITHMETIC. ) Viv 


rejecting or dropping the number 9 as often as the sum 
amounts to that ‘number, and proceeding with the excess,| 


and finally denote the last excess. Perform the same 
operation upon each of the factors ; then multiply together 
the excesses of the factors, and cast out the nines from! 
their product. If the excess of this smaller product be 
equal to the excess of the larger product first found, the 
work may be supposed to be right. It is, however, to’ 
be observed, that, although this test furnishes satisfactory 
evidence of the correctness of an operation, it is not an 
infallible proof; for, if a product chance to contain an 


error of just 9 units of any degree, the excess of its, 


horizontal sum is not thereby altered. : 
In order to perceive why the excess above nines found 


in the horizontal sum of a product, must be equal to the 


excess found in the product of the excesses of the fac- 
tors, observe that, by the law of notation, a figure is 
increased nine times its value by its removal one place 
to the left; and hence, however far a figure is removed 
from the place of units, when its nines are excluded, its 
remainder can be only itself. Therefore, any number, 
and the horizontal sum of its figures, must have equal 
remainders when their nines are excluded. This being 
understood, observe that, since factors composed of. 
entire nines will give a product consisting of entire nines, 


it follows, that any excess above nines im a product, 


must arise from an excess above nines in the factors. 
Therefore, the product of the excesses of the factors, 
must contain the same excess that is contained in the 


product of the whole factors. : 


Was 
DIVISION. 


Divisron is the operation by which we find how many 
times one number is contained in another. Itis the con- 
verse of multiplication; the product and one factor being 
given, and the other factor resultmg from the operation. 


: 
iY 


VI. DIVISION. 15 


The number which corresponds to the product in 
multiplication, is the number to be divided, and is called 
the dividend. ‘The given factor is the number to divide 
by, and is called the divisor. The factor to be found, 
that is, the number which shows how many times the 
dividend contains the divisor, is called the quotient. 

As multiplication has been shown to proceed from 
addition, so division may be shown to proceed from 
subtraction.. If we repeatedly subtract the divisor from 
the dividend till the latter is exhausted, the number of 
subtractions performed will answer to the number of 
units in the quotent. For example, if the dividend be 
24, and the divisor 6, the quotient may be found by sub- 
traction thus, 24—6—18, 18—6—12, 12—6=6, 
6—6=0. Llere 6 is subtracted four times from 24, 
and there is nothing remains; therefore,4 is the number 
of times that 6 is contained in 24. In division, this oper- 
ation is denoted thus, 24+6==4; or thus, 74 —=4. 

Division not only investigates the number of times the 


- dividend contains the divisor, but it also serves to divide 


the dividend into as:many equal parts as the divisor con- 
tains units; the quotient being one of ihese paris. ‘Tis 
effect of the operation may be understood by consider- 
ing, that, since the divisor and quoticat are factors of the 
dividend, they must each indicate how many of the other 
the dividend contains. 

It may be observed, that all the preceding operations 
begin at the place of simple units; division, however, 
must begin at the highest degree of units; for, the number 
of times that the divisor is contained in the higher units 
of the dividend must be taken out first, in order that any 
remainder, or excess above an exact number of times, 
may be carried down to the lower degrees of units, and 
divided therewith. 

When the divisor is not contained an exact number 
of times in the dividend, there will be a remainder at the 
end of the operation. This remainder, being a part of 
the dividend, is to be divided; but its quotient will be 
smaller than a unit, since a quantity in the dividend just 
equal to the divisor, gives only a unit in the quotient. 


or 


16 ARITHMETIC, VI. 


Quantities smaller than a unit, that is, parts of a unit, are’ 
called Fractions. Such quantities are commonly ex-' 
pressed by two numbers, placed one above the other. 
with a line between them, thus $. The lower number, 
called the denominator, shows how many equal parts the, 
unit is divided into; and the upper number, called the 
numerator, shows how many of the equal parts are em-. 
braced in the fraction. When the unit is divided into’ 
two equal parts, the parts are called halves; when divided. 
into three equal parts, the parts are called thirds; when 
divided into four equal parts, the parts are called fourths; 
and so on; the number of the denominator giving the 
name. For example, if the unit be divided into five equal | 
parts, one of the parts is denoted thus, 4, and called one-. 
fifth; two of the parts, thus, 2, and called two-fifths; and 
so on. In this method, the unit may be divided into any 
number of equal parts, and any number of such parts may 
be denoted. sig 
With this elementary view of fractions, it may be per- 
ceived, that when there is a remainder of 1 unit, it is to. 
be divided into as many equal parts as there are units in| 
the divisor, and one of these parts is to be annexed to 
the quotient. ‘This is performed by merely writing the 1 | 
as a numerator, and the divisor as the denominator, on’ 
the right of the quotient. If the remainder be 2 units, 
there will be 2 such parts of a unit as the divisor indicates | 
to be annexed to the quotient, and, therefore, the nume- 
rator will be 2. If the remainder be 3 units, the numera- 
tor will be 3; andsoon. Hence, whatever the remainder | 
may be, it becomes, in the quotient, the numerator of a | 
fraction, the divisor being the denominator. | 
RULE FOR DIVISION. When the divisor does not ex-— 


ceed 9, draw a line under the dividend, find how many | 
times the divisor is contained in the left hand figure, or 
two left hand figures of the dividend, and write the figure — 


expressing the number of times underneath: if there be a 


remainder over, conceive it to be prefixed to the next fig- 


ure of the dividend, and divide the neat figure as before 
Thus proceed through the dividend. 
When the divisor ts more than 9, find how many times 


‘ 


| VE, DIVISION. 17 


it ts contained in the fewest figures that will contain it, 


on the left of the dividend, write the figure expressing 
the number of times to the right of the diwidend, for the 
first quotient figure ; multiply the divisor by this figure, 
and subtract the product from the figures of the dividend 
considered. Place the next figure of the dividend on the 
right of the remainder, and divide this number as before. 
Thus proceed through the dividend. [If there be a final 
remainder, place it as a numerator, and the divisor as a 
denominator, on the right of the quotient. 

PROOF. Multiply the whole numbers of the divisor and 
quotient together, and to the product add the numerator 
of any fraction in the quotient: the sum will be equal to 
the dividend, tf the work be right. ; 

Abbreviations of the above rule may frequently be 
adopted, as follows: 

When there are ciphers on the right hand of a divisor, 
cut them off, and omit them in the operation; also cut off 
and omit the same number of figures from the right hand 
of the dividend. Finally, place the figures cut off from 
the dividend, on the right of the remainder.- 

When the divisor is 10, 100, 1000, &c.5 cut off as 
many figures from the right hand of the dividend as 
there are ciphers in the divisor; the other figures of the 
dividend will be the quotient, and the figures cut off will 
be the remainder. 

When factors of the divisor are known, diwide the 
dividend by one of these factors, and the quotient thence 
arising by the other: the last quotient will be the true 
one. To find the true remainder, multiply the last re- 
mainder by the first divisor, and to the product add the 
first remainder. 


1. Divide 4062900311 by 9, and prove the operation. 

2. How many times is 502 contaimed in 742607107 

3. Suppose 52076348 to be a dividend, and 8649 the 
divisor; what is the quotient? 

4. If 26537009535 be divided into 27856 equal parts, 
what will be one of those parts? 

5. Divide 16500269842 by 86000 ; abbreviating. 

Q* 


F 


a 


is ARITHMETIC. VIL 


6. Divide 8065743924 by 10000; by abbreviation. 
7. Divide 290516 by 63; using factors of the divisor. 
8. 142375800392 + 5274 —=what number? | 


Vil. 
PROPERTIES OF NUMBERS. 


Before proceeding to examine the properties of num-, 
bers, a few arithmetical terms, which we shall here 
collect and define, should be perfectly understood. As. 
an exercise in this article, the learner may give, upon his 
slate, an example of each term defined, and each prope 
erty described. | 

A UNIT, or UNITY, is any thing considered individual- | 
ly, without regard to the parts of which it is composed. 

An INTEGER is either a unit or an assemblage of units; } 
and a FRACTION is any part or parts of a unit. 

One number is said to MEASURE another, when it 
divides it without leaving any remainder. : 

A number which divides two or more numbers with- 
out a remainder, is called thelr cOoMMON MEASURE. 

When a number can be measured by another, the for- 
mer is called the MULTIPLE of the latter. 

If a number can be measured by two or more numbers, 
it is called their coMMON MULTIPLE. ' 

A COMPOSITE NUMBER is that which can be measured | 
by some number greater than unity. 

The auiquotT parts of a number, are the parts by’ 
which it is measured, or into which it can be divided. | 

An EVEN NUMBER is that which can be measured, or’ 
exactly divided by 2. 

An ODD NUMBER is that which cannot be measured 
by 2; it differs from an even number by 1. | 

A PRIME NUMBER is that which can only be measur- 
ed by unity, that is, by 1 ‘| 

One number is PRIME TO ANOTHER, when unity is | 
only number by which both can be measured. 


VIL PROPERTIES OF NUMBERS. 19 


A SQUARE NUMBER is the product of two equal fac- 
tors; or, the product of a number multiplied by itself. 

The sQuARE ROOT is the number, which, being mul- 
tiplied by itself, produces the square number. 

A cuBE is the product of three equal factors; or, the 
product of a number twice multiplied by itself. 

The cube rRooT is the number, which, being twice 
multiplied by itself, produces the cube. 


Property 1. The sum, or the difference of any two 
even numbers, is an even number. 

Prop. 2. The sum, or difference, of two odd num- 
bers is even; but the sum of three odd numbers is odd. 
_ Prop. 3. The sum of an even number of odd num- 
bers is even; but the sum of an odd number of odd num- 
bers is odd. 

Prop. 4. Thesum, or the difference of an even num- 
ber and an odd number, is odd. - 

Prop. 5. The product of an even, and an odd num- 
ber, or of two even numbers, is even. 

Prop. 6. An odd number cannot be divided by an 
even number, without a remainder. 

Prop. 7. A square number, or a cube number, aris- 
ing from an even root, is even. : 

Prop. 8. The product of any two odd numbers is 
an odd number. 

Prop. 9. The product of any number of odd num- 
bers 1s odd: hence the square, and the cube of an odd 
number are odd. 

Prop. 10. If an odd number measure an even num- 
ber, it will also measure the half of it. 

Prop. 11. Ifa square number be either multiplied or 
divided by a square, the product or quotient is a square. 

Prop. 12. If a square number be either multiplied 
or divided by a number that is not a square, the product 
or quotient is not a square. 

Prop. 13. The difference between an integral cube 
and its root, is always divisible by 6. 

Prop. 14. The product arising from two different 
prime numbers cannot be a square. 


2) ARITHMETIC. VIL 


Prop. 15. ‘The product of no two different numbers, 
prime to each other, can make a square, unless each of 
those numbers be a square. 

Prop. 16. Every prime number above 2, is either ] 
greater or 1 less than some multiple of 4. | 

Prop. 17. Every prime number above 3, is either | 
greater or 1 less than some multiple of 6. ‘ | 

Prop. 18. The number of prime numbers is unlimit 
ed. ‘The first ten are, 1, 2, 3, 5, 7, 11, 13, 17, 19, 23 
The learner may find the succeeding ten. 


sib emma | 
PROBLEMS. | | 


A PROBLEM is a proposition or a question requiring 
something to be done; either to mvestigate some truth 01 
property, or to perform some operation. | 

he following Problems and Rules are founded in thé) 
correspondence of the four principal operations of arid 
metic; viz. Addition, Subtraction, Multiplication, anc 
Division. 


—_- 


ii 


PROBLEM I. ‘The sum of two numbers, dag one ol 
the numbers being given, to find the other. RULE. Sub: 
tract the given number from the given sum; the remain 
der will be the number required. 

1. Suppose 37486 to be the sum of two numbers, one 
of which is 8602; what is the other? 

2. 33000 news-papers are sold im London, daily: of 
these, 17500 are morning papers, the rest, evening: how 
many of the latter? 


PROBLEM Il. ‘The difference between two numbers, 
and the greater-number being given, to find the smaller, 
RULE. Subtract the difference from the greater number; 
the ene will be the number required. . | 

If 1406 be the difference between two numbers, 
and the greater number be 4879, what is the smaller? | 


VIIL. PROBLEMS, 21 


_ 4, The area of North and South America is 18000000 
square miles: that of North America is 11000000: what 
is that of South America? 


PROBLEM II. The difference between two numbers, 
4nd the smaller number being given, to find the greater. 
RULE. Add the smaller number and the difference 
‘ogether; the sum will be the number required. 

5. Suppose 86974 to be the difference between two 
aumbers, and the smaller number to be 7064; what is 
he greater number? 

6. The British House of Lords consists of 427 mem- 
vers; the number in the House of Commons is 131 great- 
st. How many are there in the House of Commons? 


PROBLEM Iv- The sum and difference of two num- 
vers being given, to find the numbers. RULE. Sub- 
ract the difference from the sum, and divide the re 
nainder by 2; the quotient will be the smaller number. 
Then add the given difference to the smaller number, 
vid this sum will be the greater number. 
~%. What are the two numbers whose sum is 1094, and 
whose difference is 154? 3 
_§. The United States Congress, consisting of a Sen- 
ite and House of Representatives, has 288 members. 
The House has 192 members more than the Senate. 
dow many in each branch? 


_ PROBLEM V. The product of two factors, and one 
of the factors being given, to find the other. RULE. 
Divide the product by the given factor, and the quotient 
vill be the required factor. 

9. 1246038849 is the product of some two numbers, 
me of which is 269181: what is the other? 

10. Suppose a session of Congress which continues 
(80 days, to cost 504000 dollars; what is the expense 
yer day, to the United States? 


PROBLEM Vi. The dividend and quotient being given 


| 
22 ARITHMETIC. Vit 


to find the divisor. RULE. Divide the dividend by th 
given quotient, and the quotient thence arising will | 
the number sought. | 

11. Suppose 101442075 to be a dividend, and 402 
the quotient; what is the divisor ? 4 

12. 17155 pounds of beef having been equally divide 
among a number of soldiers, each one found that hj 
share was 47 pounds. What was the number of soldiers 


PROBLEM Vil. .'The divisor and quotient being giver| 
to find the dividend. RULE. Multiply the divisor an) 
quotient together ; the product will be the required aq 
idend. 

13. If 800027 be a divisor, and 97563 the quotied 
what number is the dividend ? 

14. A quantity of beef was divided equally amoal 
2742 soldiers, and each soldier received for his shar 
152 pounds. What quantity was divided ? | 


PROBLEM VIII. The product ‘of three factors, an 
two of those factors bemg given, to find the third factoy 
RULE. Find the product of the two given factors, an, 
by this number divide the given product; the quotes 
will be the factor required. 

15. Suppose the product of three factors to be 1344 
one of these factors being 12, and another 8; what is th/ 
third factor ? 

16. How many days will 9720 pounds of hay last 1| 
horses; allowimg each horse to eat 45 pounds a day ? 


PROBLEM IX. Two numbers being given, to find thet 
greatest common measure; that iS, the greatest numbe 
which will divide them both without aremainder. RULE 
Diwide the greater number by the smaller, and this di’ 
visor by the remainder, and thus continue dividing th’ 
last divisor by the last remainder, till nothing remains 
The divisor last used will be the number required. 

When the greatest common measure of more than tw 
numbers is required, first, find the greatest common mea 
sure of any two of the numbers, then find the my 


Ce 


VIII. PROBLEMS. 238 


common measure of the number found and another of the 
riven numbers, and thus proceed, till all the gwen num- 
vers are brought in. 

/ 17. What is the greatest common measure of 918, 
1998, and 522? 


918)1998(2 54) 522(9 
| 1836 — 486 
| 162)918(5 36)54(1 
| 810 la 
| 108) 162(1 18)36(2 
108 36 
| 34)108(2 
) 108 Ans. 18. 


The truth of the rule in this problem will be discovered 
oy retracing the first of the above operations, as follows. 
Since 54 [the last divisor] measures 108, it also measures 
LOS -+-54, or 162. Again, since 54 measures 108 and 
162, it also measures 5 X 162+-108, or 918. In the 
same manner it will be found to measure 2 X 918 + 162, 
yx 1998. Therefore, 54 measures both 918 and 1998. 
[tis also the greatest common measure; for, suppose there 
de a greater— then, since the greater measures 918 and 
1998, it also measures the remainder, 162; and since it 
iigasures 162 and 918, it also measures the remainder 
(08; in the same manner it will be found to measure ‘the 
‘emainder, 54; that is, the greater measures the less, 
which is absurd. 

18. What is the greatest common measure of the num- 
ers, 323 and 425? 

19. What is the greatest common measure of 2314 
ind 4626? 

_ 20. What is the greatest common measure of 1092, 
1428, 1197 and 805? 

' 21. Suppose a hall to be 154 feet long, and 55 wide; 
what is the length of the longest pole, that will exactly 
measure both the length and width of the hall ? 

| 22. A owns 720 rods of land, B owns 336 rods, and 
C 1736 rods. They agree to divide their land into equal 
house lots, fixing on the greatest number of rods for a lot, 


f 
' 


24 ARITHMETIC. Vit 


that will allow each owner to lay out all his land. Ho 
many rods must there be in a lot? Pl 

PROBLEM X. ‘Two or more numbers being given, t) 
find their least common multiple; that is, the least num 
ber that will contain each of the given numbers a whol 
number of times. RULE. Divide two or more of th 
given numbers by any prime number that will measur) 
them, repeat the operation wpon the quotients and und) 
vided numbers, and thus continue, till they become prim, 
to each other. Multiply the several divisors, the la, 
quotients, and undivided numbers together; the prod@ 
will be the least common multiple. | 

If, among the numbers to be divided, any number is 
measure of another, the measuring number may he | 
jected; that is, dropped from the operation. 

It is obvious, that one number is the multiple of anothal 
when the former contains all the factors of the latte). 
The factors of 6 are 3 and 2, and the factors of 9 are 
and 3. Now 54 contains all these factors, (3 x 2 3) 
3== 54), and 54 is a common multiple of 6 and 9, but | 
is not their least common multiple—it is 3 times as gre: 
as the least, owing to the existence of the factor, 3,1 
both 6 and 9. Hence we observe, that a common facto 
of two or more numbers must enter but once into th 
multiplication, to give the least common multiple. Th 
above rule effects the necessary exclusion. 

23. What is the least common multiple of 12, 25, at) 
and 45. 

3) 12 25 30 45 We find, after dividin\ 
by4eles 210N15 . | twice, that 4 and 2 ap! 
——____ | pear; and, . by» droppigy 
4 5 2 38 |the 2 because it measure, 
the 4, we avoid angie 

3X5X4xX5 xk 3=900 division. ins. 900. | 


24. What is the least common multiple of 6, 10, 6 
and 20? 

25. What is the least common multiple of 25, 35, 60) 
and 72? | 


; 
| x: COMPOUND NUMBERS. 25 


and 245? 
27. What is the least common multiple of 18, 82, 94, 
788, and 356? 
98. Allowing 63 gallons to fill a hogshead, 42a tierce, 
‘and 32 a barrel, what is the smallest quantity of molasses, 
that can be first shipped in some number of full hogs- 
| heads, then discharged and reshipped in some number of 
‘full tierces, and again Biaghasead and reshipped in some 
number of full barrels ? 
/ ~29. Acertain flour dealer, who purchased his flour 
from a mill on the opposite side of a river, owned four 
_ boats, one of which would carry 8 barrels of flour, another 
9, another 15, and another 16. What is the smallest 
“number of barrels he could purchase, that would make 
- some number of full freights for either of the boats ? 


i - 26. What is the least common multiple of 105, 140, 
: 
' 


| IX. 
y Pd 7 
! COMPOUND NUMBERS. 


Compounp Numsers are those‘which are employed 
_to express quantities that consist of several denominations; 
-each denomination being denoted separately. Under this 
Bead are classed, all the subdivisions of measures; of 
‘Yength, surface, solidity, weights, money, time, &c. 

The following tables of denominations of compound 
_numbers, show how many units of each lower denomina- 
_tion are equal to a unit of the next higher, and, exhibit 
each lower denomination as a fraction of the next higher. 


MONEY, WEIGHTS, AND MEASURES. 


ENGLISH MONEY. 

The denominations of English Money are, the pound, 
_£, the shilling, s., the penny, d., and the farthing, qr. 

4 farthings Bes Fi = 1d. | Lar. (ie eee ge Le 
a a =1s. <a cena 
peelings... abba 1S. ss ae, aoe Oe, Ld 

4 a 


v 


26 ARITHMETIC. 1X: 


TROY WEIGHT. | | 
The denominations of Troy Weight are, the pound, Ib., 
the ounce, oz., the pennyweight, dwt., and the grain, gr. 


DA Vai a tes ss =e —Idwt.| lgr. ...—=,o0f 1dwt.. 
20 pennyweights ...==1loz. | ldwt A: of 1oz 
12Q%euMEes HOU == 1 lb. loz. ..== q0f 1b. 


AVOIRDUPOIS WEIGHT. 
The denominations of Avoirdupois Weight are, the ton, 
T., the hundred-weight, cwt., the quarter, gr. the pound, 
lb., the ounce, 0z., and the dvand, dr. 


16idralls Adenia A —loz. {| 1dr. Sas ahs 
16°ountes sos: ==I1lb. | loz. ...—-,,0f 1b 
2S poundsh ss. 5 ee —-1qr. || ilb. — x5 of 1 qr. 

4 qiiarters «45. 5... = lewt.| Iqr. ... = i of lewtt) 
20 hundred-weight .—=-1T. || lewt...=j,of 1T. | 


APOTHECARIES’ WEIGHT. 
The denominations of Apothecaries’ Weight are, the. 
pound, fb, the ounce, 3, the dram, 3, the scruple, Dy 
and the grain, gr. 


2Ograins i.) .).4 eee a1 DEY Vere oi eee of 1D. 
2 scruples. ......% 6215. 150 ..... = 27 of 1501 
Gidea. wee eee d Soul ns UA, — 1 of 13a 
12 ounces 25.4.7; bibs dS. oo. eee tes | 


CLOTH MEASURE. 
The denominations of Cloth Measure are, the French 
ell, Fr. ¢., the English ell, E. e., the Flemish ell, Fl.e.,| 
the yard, yd., the quarter, Gr; and the nail, na. 


A TWAS “Ge ee I qr. ; Ina. att of 1 qr. 
ANjiarters 2.53 - >: eee Dy. OW Pgh. 05 eet tae es | 
3 quarters ....... a= 1 Fle.) Vor.. oS eer ae | 
S.qQuarters s:.1 44> —I1E. e.| | I qr. lof 1E. ef 
5 quarters ....... a=) Fre. 1 qr’ Soe oF ee ‘ 


DRY MEASURE. 

The denominations of Dry Measure are, the bushel, 

bu., the peck, pk., the gallon, gal., the quart, gi., and 
By pint, pt. 


PT i a re ==Iqt. | 1 pt. ee | 
4 quarts ...... pate Leal Varts :... 222 of Leal 
Siquartsa he. 0. .ae = 1p I qt WD 
WOCKA eh ss 35s = Ibu. |} Ipk. ....—=4o0f lbu 


' COMPOUND NUMBERS. a7 


‘ WINE MEASURE. 

The denominations of Wine Measure are, the ton, 7., 
the pipe, p., the puncheon, pun., the hogshead, hhd., the © 
tierce, tier., the barrel, b/., the gallon, gal., the quart, 

gt. the pint, pt. and the gill, gi. 


Met cills . .:.... sail pte tibeiwi. com cohhot 
lmmepints ....... == 1 qt. Ppt a Sof qh 
wet quarts... ... saul galviit hgts syed oofkieal 
| B13 gallons ..... — 1bl. lgal... = 4 of 1bl. 

me gallons). . ...- = Itier. | 1gal... = 4, of I tier. 
| 63 gallons ...... == [hhd. | lgal... = 4 of 1hhd 
| 84 gallons ...... — pun. | Igal... = 4 of 1 pun. 

mo gallons .:..... —Ip. 1 Bal Seer oleh ty 
PIPES: Hore se LPuet hp asete tof PT: 


BEER MEASURE. 
The denominations of Beer Measure are, the butt,,bé:, 
the hogshead, hhd., the barrel, bl.,the kilderkin, kil., the 
firkin, fir., the gallon, gal., the quart, gt., and the pint, pt. 


REOtGe o 2c ee == iat. | lpt. ... = 4 of 1lqt. 
BMAENS © ci wie 83 == i galon) Pgteri.: == 4 ofiligal: 
Gallons ....... == fir. | Igal....-= 4 of I fir. 
Seorkins ........ == [Tkil. | Ifir. ».. = 4 of 1kil. 
Qkilderkins ..... = Ibl. Blea. Be 4 of 1bl. 
3 kilderkins ..... =="Thhds Pelkih 2.9 4) of Lbhd. 
2hogsheads ..... —I1bt. | Lhhd: .. — 3 of 1bt. 


“NOTE. In the United States, the Dry gallon contains 
2684 cubic inches, the Wine gallon 231 cubic inches, 
and the Beer gallon 182 cubic inches. By an Actof the 
British government, however, the distinction between 
the Dry, Wine, and Beer gallon was abolished in Great 
Britain, in 1826, and an Imperial Gallon was established, 
as well for liquids as for dry substances. The Imperial 
gallon must contain ‘10 pounds, Avoirdupois weight, of 
distilled water, weighed in air, at the temperature of 62° 
, of Fahrenheit’s thermometer, the barometer standing at 
30 inches.” This quantity of water will be found to 
measure 2772/4 cubic inches. ‘The same Act estab- 
lishes the pound Troy at 5760 grains, and the pound 
_ Avoirdupois at 7000 grains. 


28 ‘ARITHMETIC. IX. 


LONG MEASURE. 

The denominations of Long Measure are, the mile, Mis 

~ the furlong, fur., the rod or ‘pole, r., the yard, yd., the 
foot, ft., and the inch, in. 


12‘inchessa vy 3. == 1] fee JL mis os eee of ft 
SBOE gs pW ni dacs == lyd. || 1ft. = 3 of Lyd 
RaROOS 414-3 5h = Ir. lyd.... = +4, of In@ 

40 SOEs yb cette ts eo Adu fdr. oS. = qo of 1 fur 
8 furlongs... -.. — 1m. |) 1 for. = + of lm 


SQUARE MEASURE. 

The superficial contents of any figure having four sides 
and four equal angles, is found in squares, by multiplying | 
together the length and breadth of the figure. 4j 
The denominations of Square Measure are, the mile, | 
mm., the acre, A., the rood, R., the rod, r., the yard, yd. i 
the foot, ft., and the inch, in. | 
144 inches....... 21 ft. y Lin. .. == qhy of I ft. 


CUBIC MEASURE. 
The cubical contents of any thing which has 6 sides— 
its opposite sides being equal—is found in cubes, by) 
multiplying together, the length, breadth and depth. 
The deaaininations of Cubic Measure are, the yard, yd., 
the foot, ft., and the inch, in. 

1728 inches. cit vues) Bi ea ee = ray Of I ft. 
ya fe Tc nee = 1yd. | 1ft. .. = 3 oflyd. 
40 feet of round timber, or 50 feet of hewn timber 

make a ton. 16 cubic feet make a foot of wood, and 8) 

feet of wood make a cord. 

TIME. 
The denominations of Time are, the year, Y., the day, | 

d., the hour, h., the minute, m., aod the second, 5 


fi = 7 
9 feeb shims bh =< Lyd. ft. ..o== dof hyd 
304 yards>...... = ir, lyd == 737 Of Ir | 
40:70dS -¢.5eSde. —I1R. | 1r......= gy of URE 
A POOKS is ark x Ji A. 4 LR: == of iol 
640 acres yee oe ew Aba AAS = g}y of Im. | 


— 


i 


| 
1 
f 
} 


60 seconds ....., 1m. / 1s. rire ae ce | 
60 minutes ...... — Ih. | lm = 7, of th 
DAT OUTE IS fais. « dtd = 1d: | Lhe. = 5, of ld 
BOD MEY Voie. > 8 == 1 Y, lia legs’ ik = zt, of LY 


1X. COMPOUND NUMBERS. 29 


The earth revolves round the sun once in 365 days, . 
5 hours, 48 minutes, 48 seconds: this period is therefore — 


a Solar year. In order to keep pace with the solar year, 


in our reckoning, we make every fourth year to contain 
366 days, and call it Leap year. Still greater accuracy 
requires, however, that the Leap day be dispensed with 
3 times, in every 400 years. Whenever the number 
which denotes the year can be measured by 4, the year 
is Leap year— the centurial years excepted. 

The year is also divided into 12 months—See Almanac. 

THE CIRCLE. 
The divisions of the circle, C., are, the sign, S., the 


_ degree, (°), the minute, (’), the second, (”). 


ee 


This table is applied to the Zodiac; and by it are com- 
puted, planetary motions, latitude, longitude, &c. 


mOseconds « .. «..’. — j! | Lik’ shes he == gy of 1° 
Se mUteS) <. 6.» = ry: } Wnts tie == gy of 1° 
30 degrees ....... — i i2 . == 3 of 1S. 
12 signs ee ‘aati | is. == zy Of IC. 


GEOGRAPHICAL MEASURE. 
- The circumference of the globe—like every other cir- 


cle—is divided in 360 equal parts, called degrees. Hach 


degree is divided mto 60 equal parts called miles, or 
minutes. Three miles are called a league. 

On the equator, 69+ statute miles are equal to 60 geo- 
graphical miles, or | degree, nearly: and, on the meridian, 
at a mean, 69); statute miles are equal to a degree. 


REDUCTION OF COMPOUND NUMBERS. 


Repvucrion is the operation of changing any quantity 
from its number in one denomination, to its number in 
another denomination. } 

RULE FOR REDUCTION. When a greater denomina- 
tion is to be reduced to a smaller, multiply the greater 
denomination, by that number which is required of the 
smaller, to make a wnit of the greater; adding to the 


_ product, so many of the smaller denomination as are ex- 


pressed in the given quantity. Perform a like operation 
on this product, and on each succeeding product. 
* 


. 


30 ARITHMETIC. 1X. 


When @ smaller denomination is to be reduced to a 
greater, divide the smaller denomination by that number 
which is required of the smaller, to make a unit of the 
next greater: the quotient will be of the greater denomi-_ 
nation, and the remainder will be of the same denomina- 
tion with the dividend. Performa like operation on this: | 
quotient, and on each succeeding quotient. © : 


. Reduce £351 13s. Od. 1 qr. to its value in farthings. | 
. How many pounds, &c. are there in 6169 pence > | 
In 591b. 13dwt. 5gr. Troy, how many grains? ~~ 
Change 20571005 drams to its value in tons, &c. 
In 231ib 33 05 0D 5gr. how many grains ’ 
. How many English ells are there in 352 nails ? 
. Reduce 7 bushels and 6 quarts to pints. 
. How many hhds. are there in 9576 pints of wine? . 
9. How many pints im 1 bl. 1 fir. 1 pt. of beer ? 7 
~ 10. How many miles, &c. are there in 26431 rods? 
11. In 3 square miles, how many square rods ? yy 
12. In 1259712 cubic inches, how many cubic yards ? } 
13. Reduce 1 solar year, 7d. and 10h. to seconds. 


cag tree eee a 


ADDITION OF COMPOUND NUMBERS. 


The operation of adding compound numbers, differs 
from that of adding simple numbers, only, with respect | 
to the irregular system of units, which determines the | 
principles of carrying from one denomimation to another. | 

RULE. Write the numbers so that each denomination | 
shall stand in a separate column. Add the numbers of | 
the lowest denomination together, and divide their sum | 
by that number which is required of this denomination to | 
make @ unit of the next higher: write the remainder un- | 
der the column added, and carry the quotient to the next | 
column. Thus proceed through the denomination. 


14. What is the sum of £9 8s. 4d., £250 8s. 5d. 
3qr., £9 7s. 4d., £20 16s. 4d., and 3s. 6d. 2qr. 7 | 
15. Add together 100z..14dwt. 16gr., 5lb. 9oz. | 
6dwt. 22¢r., 4lb. Loz. 18 awt. Qgr., and 11 dwt., Troy | 
16. Add together 15T. 19cwt. 3qr. 21b. 702., 25 | 
13cwt. 2qr. 20lb. 150z., and 3qr. 26]b. 


x. COMPOUND NUMBERS. ot 


_ 17. How much is 18 yd. 3qr. 3na., Loyd. 2qr. 3na., 
%yd. lqr. 2na., and 57yd. 3qr. 2na. of cloth? 

18. Add together 25bu. 3pk. 7qt., 100bu. 2pk. 4qt., 
’115bu. 2pk. 2qt. lpt., and 57bu. 3pk. of corn. 

19. Add together 4p. 125gal. 3qt., 75gal. Qqt. 1 pt., 
35p. 92gal., and 39 gal. 3qt. 1 pt. of wine. , 
_ 20. How many acres are 13A. 3R. 38r., 87A.2R. 
33r., 26A. 2R., 41 A. 2R. 28r., and 36r.? ‘ 
| 21. How much hewn timber is 9T. 19ft. 1725in., 
150T. 39ft. 1695in., and 500T. 31 ft. 915in.? 


SUBTRACTION OF COMPOUND NUMBERS. 


RULE. Wriie the several denominations of the smaller 
juantity under the same denominations of the greater 
juantity: then, begin with the lowest denomination, and 
verform subtraction on each denomination separately. 
Whenever a number expressing a denomination in the 
spper line is smaller than the number under it, increase 
he upper number by as many as make a unit of the next 
ugher denomination, and consider the number of the 
vext higher denomination in the upper line, to be 1 less 
han tt stands. 


22. Subtract ilb. 100z. 16dwt. from 3lb., Troy. 
23. From 6T. 3cwt. take 7cewt. 2qr.15lb., Avoir. 
24. From 2th 73 take 73 63 2D 5er., Apoth. wt. 
25. Subtract 3qr. 3na. from 5yd. 2qr. Lna. of cloth. 
26. Subtract 8bu. 1 pk. 6qt. 1 pt. from 50bu. of corn. 
27. From 3hhd. 25gal. take 41 gal. 2qt. of wine. 
28. From 6bl. 1kil. take 1 fir. 6 gal. Sqt. of beer. 
29. Subtract 3yd. 10in. from 5yd. 2 ft. oi Todas 
30. Subtract 57 A. 2R. 31r. fromim., Square mea 
. 31. Subtract 2Y. 90d. 4h. 55m. from 4Y., Time. 


» 
a 


MULTIPLICATION OF COMPOUND NUMBERS. 


RULE. Begin with the lowest denomination, and mul- 
‘ply each denomination separately; divide each product 
by the number which is required of its own denomination 


| 


} 


ea 


32 ARITHMETIC. 


io make a unit of the next higher; write the remainde 
under the denomination multiplied, and carry the quotiel 
to the product of the next higher denomination. | 
| 
| 


32. Multiply £215 19s. 6d. by 72 or its factors. 
33. Multiply 2lb. 50z. 7dwt. 10gr., Troy, by 56. 
34. What is 16 times 18cwt. 3qt. 15]b., 1402.? 
35. What is 81 times 36bu. 3pk. 6qt. 1 pt., Dry me: 
36. Multiply 4p. 105gal. 3qt. of wme by 60. 
37. Multiply 2m. 7fur. 35r., Long mea., by 63. | 
38. Multiply 4m. 320A. 1R. 9r., Square mea., by 15 
39. Multiply 2Y. 250d. 14h. 30m., Time, by 96. | 
| 


% 
x 


DIVISION OF COMPOUND NUMBERS. | 

RULE. Divide each denomination separately, begin 
ning with the highest. Whenever a remainder occurs 
reduce it to the next lower denomination, add it to th 
number expressed in the lower denomination, and divid’ 
it therewith. | 
40. Divide £251 15s. 7d. 2qr. into 46 equal parts.) 
41. Divide 15lb. Soz. 7dwt. 5gr., Troy, by 13. 
42. Divide 12T. 27 lb. 150z., Avoirdupois, by 5. 
43. Divide 136 K.e. 3qr. 3na. of cloth by 31. | 
44. Divide 1621 bu. 2pk. of corn into 50 equal parts’ 
45. Divide 1 pipe of wine equally among 9 owners. | 
46. Divide a Leap year into 100 equal paris. | 
| 


FEDERAL MONEY. | 


f 


Ve | 


The denominations of Federal Money are, the eagle 
the dollar, the dime, the cent, and the mill. 16 mill’ 
make 1 cent, 10 cents 1 dime, 10 dimes 1 dollar, and 1! 
dollars 1 eagle. Dollars, ¢, and Cents, ets. are the onl 
denominations commonly mentioned in business— eagle) 
being counted as tens of dollars, dimes being counted a 
tens of cents, and mills not being denoted. j 
100 cents ........ == $1 || Lcent... =), of $i] 


The cents in any number of dollars are expressed by 
the same figures which express the dollars, with twe 


! 
IX. COMPOUND NUMBERS. 33 


siphers annexed; $15==1500 cents. The dollars in any 
number of cents are distinguished by cutting off two fig- 
wes from the right for cents; 325 cts. — $3.25. 

Operations on numbers expressing Federal money, are 
yerformed as on simple numbers; care must however be 
aken, in addition and subtraction, to place dollars under 
lollars, and cents under cents; these denominations being 
veparated by a point. 


47. What is the sum of $34.21, $7064.04, 36cts., 
$10004.85, $96, $900.10, $14, $1.99, and #76529 ° 
"48. Subtract $4926 from $ 12262.37. 

49. Subtract $297.18 from $ 100000. 

50. Suppose $295.48 to be a multiplicand, and 25 the 

nultiplier; what is the product? 

_ In multiplication, only one of the factors can be Federal 
oney, and the product will be of the same denomination 
js this factor. If, therefore, there be cents in either fac- 
jor, two figures must be pointed off for cents, from the 
ight of the product. 
_ 51. What is the product of 96 cts. multiplied by 43 ? 
_ §2. What is the value of 1304 pounds of coffee at 9 
vents per pound ? 
_ 53. How many times $7 are there in $29.46? 
, In division, when both the dividend and divisor are 
“ederal money, they must both be of the same denomi- 
ation. If therefore, one of the numbers contain cents, 
nd the other dollars only, the latter number must have 
wo ciphers annexed to it. 

54. How many barrels of Lines at $4. 36 per barrel, 
van be purchased. for $4370? 
| 55. Divide $4279.50 into 746 equal parts. 
| 56. If 407 pounds of Hyson tea cost $395, what is 
fe cost of 1 pound? 

_ 57. How many times are 95cts. contained in $56? 
_ 58. A merchant sold 1248 yards of cloth, at such price 

s to gain 1 cent on every nail. How much did he gain > 
[ 59. What is the gain on a hogshead of molasses, sold 
t an advance of 3 cents per gallon? 
60. A jeweller sold a silver pitcher 3lb. 80z. 16dwt., 
jt 7 cents a pennyweight. What did it amount to ? 


34 ARITHMETIC. Ix 


61. What is the freight of 60480 pounds of cotton fron 
Charleston to Liverpool, at $4 per ton ? y 


MISCELLANEOUS EXAMPLES. 


62. How many bottles, holding 1 pint and 2 gills eack 
are required for bottling 4 barrels of cider? 
63. How much will 46 bushels of oats cost, at 4 pene 
2 farthings for every two quarts ? | 

64. A brewer sold 96 hogsheads of beer for £38) 
16s. What was the price of 1 pint at the same rate ? 

65. A certain tippler spent 12 cents a day for arder| 
spirit, during 39 successive weeks, and then died, the vi 
tim of his folly. What did the spirit all cost ? 

66. Bought five loads of wood; the first containing — 
cord 32 cubic feet, the second 1 ee 64 cubic feet, th, 
third 112 cubic fect, the fourth 1 cord 28 cubic feet, ani 
the fifth 1 cord 20 cubic feet. How many cords wer 
there in the whole ? 

67. Bought goods to the amount of £25 13s. 10d) 
2qr.; and afterwards sold goods to the same man, amount. 
ing to £30 10s. 4d. 2qr. What is the balance of mone) 
in my favor? \ 

68. A farmer sold five lots of land, at $9 an acre; thi 
first lot contaming 30A. 2R. 2r., the second 41A. 3B 
8r., the third 14A.1R. 10r., the fourth 25 A. 36r. z ant 
the fifth 54A. 6r. What did the whole amount to? — 

69. How many cubic inches in a brick 8 inches long) 
4 inches wide, and 2 inches thick ? 

70. How many cubic inches in the cube of 2 inches 
meni in the cube of 3 inches?..... in the cube of ? 
inches?..... in the cube of 5 inches? 

71. If the cube of 4 inches be taken from the cube ol 
L foot, how:many cubic inches will remain ? 

72. If the cube of 4 inches be taken from the cubs 
of 2 feet, how many cubic inches will remain ? 

. 73. A young man, on commencing business, was worth 

£643 10s.; the first-year he cleared £54 11s. 7d. 2 qr. 
the second year, £87 Os. 10d. Iqr.; but the third year he 
lost £196 7s. 11d.3qr. How much was he then worth: 


oa 
ee 


| 
| 


q 


LX. COMPOUND NUMBERS. 35 


_ 74. A gentleman had a hogshead of wine in his cellar, 
rom which there leaked out 17gal. 3qt. lpt. How 
nuch then remained ? 

75. Aman started ona journey of 20 miles 6 fur. 29r., 
nd stopped to rest at a house, 4m. 4 fur. 20r. from the 
lace of starting. How far had he still to go? 

_ 76. Ina pile of wood, 96 feet long, 5 feet high, and 4 
eet wide, how many cords? 
| 77. How much would 13 hogsheads of sugar cost, at 
} cents per pounss allowing each hogshead to contain, 
sewt. 3qr. 24]b.? 
| 78. Acent weighs 8 peaniweignts 16 grains. What 
3 the weight of 100 cents? 

79. How many yards of cloth are there in 19 pieces; 
ach piece containing 27 yd. 3qr. 2na.? 

80. If a man sell 2b]. 1 kil. 1 fir. 6 gal. Qqt. 1 pt. of 
eer in #6 week, how many barrels would he sell m 
6 weeks ° 

81. If 1 pint and 3 gills of wine will fill a bottle, how 
wich will fill a gross, or 12 dozen bottles ? 

i 82. A father left an estate worth £5719 17s., to be 
ivided equally among 11 children. How much was each 
‘ne ’s share ? 

83. Sixteen men own 24 tierces of molasses, in equal 
lie. What is one man’s share ? 

84. A company of 23 men bought 1850 acres 10 rods 
‘f wild land, and divided it equally among them. How 
such land bad each mah ? 

85. What must be the length of a lot of land, that is & 
ods wide, in order that the lot shall contain 1 acre? 
| Observe in the above question, that 1 acre contains 
60 square rods; and, that this number of square rods is 
ae product of the two factors that denote the width and 
lie of the lot. See Prosiem V, page 21. 

_ 86. What must be the depth of a house lot, that mea- 
“ures 72 feet on the front, to contain 9432 square feet ? 

_ 87. What must be the length of a stick of hewn timber, 
nat is 10 inches wide and 1 ft. 3in. deep, in order that 
he stick shall contain 1 ton? 
| Observe in this question, that the number of cubic 


i 


k 
le 


36 ARITHMETIC. x 


inches in a ton, is the product of the three factors whic] 
denote, in inches, the width and depth and length of th 
stick. See ProgLemM VIII, page 22. 

88. What must be the length of a pile of wood that j 
4 feet wide and 3 feet high, in order that the pile sha 
contain 1 cord, that is, 128 cubic feet? 

89. Suppose a pile of wood to be 11 feet long and‘ 
feet wide; how high must it be, to contain 2 cords 4 fee 
of wood and 10 cubic feet ° | 


x | 
FRACTIONS. 


A FRACTION signifies one or more of the equal part 
mto which a unit, or some quantity considered as an in 
ieger, or whole, is divided. 

A fraction is expressed by two numbers or termi 
written one above the other, thus, ¢. The lower tem 
—called the denominator — denotes the number of equi 
parts into which the integer is divided; and the uppe' 
term—called the numerator—indicates what numbe| 
of those equal parts the fraction expresses. 

We may not only consider a fraction as a certain num 
ber of parts of a unit, but, may also view it as a x | 
of a certain number of units. Thus, 3 may either b/ 
considered as 2-thirds of 1, or, 1-third of 2; for 1-thir 
of 2 is the same quantity as 2-thirds of 1. Hence, if th 
numerator of a fraction be viewed as an integer, an 
divided into as many equal parts as the denominator 4 
dicates, the fraction may be regarded as expressing of! 
of these parts. Thus, if 4 be divided into 5 equal party 
the fraction ¢ expresses one of these parts. 

Fractions ‘generally have their origin from the divisia 
of a number by another which does ‘not measure it; th 
excess. of the dividend, above what can be measured b 
the divisor, being the numerator, and the divisor bem 
the denominator, as shown in Art. VI. 

If the numerator of a fraction be made equal to th 


} 
| 


, 
! 
’ 


x | FRACTIONS. 37 


} 


denominator, the fraction becomes equal to unity; thus 


$=1. If the numerator be greater than the denominator, 
the fraction is equal to as many units as the denominator 
is contained times in the numerator ; for example P= 3. 
‘Hence, a fraction may be viewed as an unexecuted divi- 
sion; the divisor being written under the dividend. It 
follows, also, that since any number divided by 1 gives 
the same number in the quotient, any number may be 
‘expressed as a fraction by making 1 its denominator. 
For example, 17 may be expressed thus, 4’. 

' The following propositions concerning fractions, should 
be distinctly noticed. 


PROPOSITION I. As many times as the numerator is 

made greater, so many times the fraction is made greater; 
and, as many times as the numerator is made smaller, so 
many times the fraction is made smaller. Hence, a jrac- 
tion is multiplied by multiplying the numerator, and 
divided by dividing the numerator. 
_ PROPOSITION Il. 4s many times as the denominator is 
made greater, so many times the fraction is made smaller ; 
and as many times as the denominator is made smaller, 
so many times the fraction is made greater. Hence, « 
fraction is divided by multiplying the denominator, and 
multiplied by dividing the denominator. 

PROPOSITION II. Whenthe numerator and denomina- 
tor are both multiplied, or both divided by the same num- 
ber, the quantity expressed by the fraction is not thereby 
changed. 
| A PROPER FRACTION is a fraction whose numerator 
us less than its denominator; as 35. 
|; An IMPROPER FRACTION isa fraction whose numerater 
equals, or exceeds its denominator; as 4, }. 
_Anumber consisting of an integer with a fraction an- 
mexed, as 142, is called a MIXED NUMBER. 
| A coMPOUND FRACTION is a fraction of a fraction; as 

of 3. 4 of 45 of 2. 

A COMPLEX FRACTION is that which has a fraction 
either in its numerator, or in its denominator, or in both 


. at 
of them; thus, 6 Ox a 7 
) 4 


* i 


| 


38 ARITHMETIC. Dy 


| 
REDUCTION OF FRACTIONS. .| 
| 
REDUCTION OF FRACTIONS consists in changing them) 
from one form to another, without altering their value. | 


CASEI. To reduce a fraction to its lowest terms; 
that is, to change the denominator and numerator to the 
smallest numbers that will express the same quantity. 

RULE. Divide both terms of the fraction by thew 
greatest common measure, and the two quotients will be the 
lowest terms of the fr action. See Pros. IX, page 22. | 
_. When the greatest common measure is readily per: 
ceived, the fraction may be reduced mentally. For in- 
stance, tne greatest common measure of the terms of the 
fraction 75, 1s 4, and the only notation necessary in the 
reduction, is, 7 =. | 

Dividing the terms of a fraction by a common measure} 
which is not the least, will reduce it in some degree, and 
when thus reduced, it may be reduced still lower by 
another division, and so on, till no Bela be will measure 
both the terms. For example, to reduce 33, divide by 2, 
and the result is 34; again, divide by 3, i and the result is) 
3. Here the fraction is known to be in its lowest terms,| 
because the terms are prime to each other. ' 

1. Reduce 84 to its lowest terms, by repeatedly 
dividing the Ae by any common measure. 

2. Reduce +55 to its lowest terms, by dividing the, 
terms by their greatest common measure. 


3. Reduce each of the following fractions to its lowell 
32 270 384 156 720 3108 


terms. Zoo: 306° isd’ 336° 736° 3559" 


a 


CASE Il. To reduce a whole number to an imprope) 
fraction. : | 

RULE. Multiply the whole number by the proposes 
denominator, and the product will be the numerator. 

When the quantity to be reduced is a miaed number: 
the numerator of the fraction in the mixed number mus) 
be added to the product of the whole number, and thei’ 
sum will be the numerator of the improper fraction. 


&. FRACTIONS. 39 


4. Reduce 16 to a fraction whose denominator is 9. 
16 In 1 unit there are 9-ninths; 
9 Seema se 9 times as 
arr Many ninths as there are units in 
144 Ans. +44 any vee ite 
Reduce 75 to a fraction whose denominator is 13. 
Reduce 3 to a fraction whose denominator is 342. 
How many fifteenths are there in 74? 
How many eighths of a dollar in $647? 
Reduce 36% to an improper fraction. 
) 364 In this example, we add the 
i 3-sevenths to the sevenths pro-. 
— ,., | duced by the multiplication of 36. 
256 Ans. 27° | by 7, and thus obtain 238 
10. Reduce 254% to an improper fraction. 
11. Reduce 61552, to an improper fraction. _ 
12. How many sixteenths of a dollar in $5412? 


eT 


CASE Il. To reduce an improper fraction to a whole 
‘number, or a mixed number. 
RULE. Divide the numerator by the denominator, and 
the quotient will be the whole, or mized number. 
13. Reduce 38? to a whole, or mixed number. 
8)362 Since $ are equal to 1 unit, 
oe there are as many units in 282 as 
there are times 8 in 362. 
14. Reduce #393 to a whole, or mixed number. 
15. How many units are there in #°31° ? 
16. How many dollars in 29? of a dollar? 


CASE Iv. To reduce a compound fraction to a sim- 
ple, or single fraction. 
RULE. « Multiply all the numerators together for a new 
numerator, and all the denominators for a new denomi- 
| nator: then reduce the new fraction to its lowest terms. 
| When any numerator is equal to any denominator, the 
operation may be abbreviated by rejecting both. 
| If part of the compound fraction be an integer, or a 
mixed number, it must first be reduced to an improper 
| fraction. : . 


oD 


40 ARITHMETIC. X! 


. Reduce ¢ of 2 of # of 6 to a sunple fraction. 
Here the common term, 3, 1s | 
Baba =33=8 | omitted in the multiplication. | 
18. Reduce 2 of 4 toa Simple fraction. ) 
9. Reduce + of + 
i) 
f 


O 

of 32 ta a simple fraction. | 
20. Reduce 44 f Soir of 2-7; to a simple fraction. f 
21. Reduce 50 of 5 toa simple fraction. | 


€ASE V. To reduce a fraction from one denomination | 
to another. i 
RULE. Multiply the proposed denominator by the 
numerator of the given fraction, and divide the product 
by the denominator of the given fraction; the quotient | 
“ee be the numerator of the proposed denominator. | 
. Reduce 2 to afraction whose denominator shall be | 
a an in other words change 5-sixths to omaaes | 
14 g is equal to % of 77, and 2 is’ 

5 5 times as mine a therefordl| 
6)70 , | find 5 times 14-fourteenths and 

— fins. 11¢ | take ¢ of this product for the | 
11% 14 | required fourteenths. j 
23. How many fifths are there in 2? 1 
24. 3% is equal to how many twenty-fourths ? | 
25. Hahn $ to a fraction whose denominator is 4. | 
26. How many twelfths of 1 shilling in + of 1s.? | 


=i 


CASE VI. To reduce the lower denominations of a 
compound number to the fraction of a higher denomination. } 

RULE. Reduce the given quantity to ; the lowest denomi- 
nation mentioned, and this number will be the numerator: | 
ihen reduce a unit of the higher denomination to the same | 
denomination with the numerator, and this number wild 
be the denominator. | 

27. Reduce 7oz. 18dwt. 13gr. to the fraction of a | 

ound. | 

We find, that 7o0z. 18 dwt. 13gr. when reduced to 
grains, gives 3805 for the numerator; and 1 pound when 
reduced to grains, gives 5760 for the denominator. 
Therefore, $422 — +%4 is the fraction required. 

28. Reduce 4s. 9d. Sqr. to the fraction of £1. 


ie FRACTIONS. 41 


29. Reduce 34 inches to the fraction of a yard. 
30. What fraction of a hogshead is 9 gal. 22 pt.? 
31. Reduce 5cwt. 8lb. 40z. to the fraction of a ton. 


CASE Vil. ‘To reduce the fraction of a higher denomi- 
nation to its value in whole numbers of lower denomination. 
RULE. JWultiply the numerator by that number of the 
next lower denomination which is required. to make a wnit 
of the higher, and divide the product by the denominator; 
the quotient will be awhole number of the lower denomi- 
nation, and the remainder will be the numerator of a frac- 
tion. Proceed with this fraction as before, and so on. 

It will be readily perceived, that the fraction of a higher 
denomination is reduced to the fraction of a lower, by 
multiplying the numerator by the number of units of the 
lower, required to make a unit of the higher. ‘Thus, 3 of 
a bushel is 4 times as many fifths of a peck; that is, of 
apeck. Again, 7 of a peck is 8 times 12-fifths, that 
is, %° of a quart; and again, %° of a quart is 2 times 
96-fifths, that is, 422 of a pint. If the denominator be 
multiplied, instead of the numerator, the effect is the re- 
verse, and the fraction is reduced to a higher denomination. 
Thus, 2 of a pint, (the 5 being multiplied by 2,) becomes 
7a of a quart; 3%; of a quart, (the 10 being multiplied by 
8,) becomes = of a peck; and ,%; of a peck, (the 80 
being multiplied by 4,) becomes 335 of a bushel. 

32. Reduce 43 of a gallon to its value in quarts, &c. 


11 We find by multiplication, that 

A +3 Of a gallon is 4% of a quart; 

i 2)44 and, by division, that 73 of a 
ph quart is 3qt. and 4% of a quart. 

3. 8 We then find, that 58, of aqt. is 

2 46 of a pint; and, that 4§ of a pt. 

12)16 is 1 pt. and +4 of a pt. And thus, 


by finding the units of one de- 


: ; nomination at a time, we finally 
—— obtain the whole answer, which, 
12)16 denoted as a compound number, 


144—14 |is 3qt. I pt. 13g). 
3. Reduce 2 of £1 to its value in shillings &c. 
4% 


¥ 

e 
~~ 
Bc 


42>, ARITHMETIC. | 


34. Reduce 4 + of a yard, to its value in feet, &c. | 
35. In 75 of Tow. how many quarters, pounds, &c.? 
36. Reduce 33 of a bushel to pecks, quarts, and pints. | 


CASE VII. To reduce fractions toa common denomi-( 
nator; thatis, to change two or more fractions which have | 
different denominators, to equivalent fractions, that shall 
have the same denominator. | 

RULE ist. Multiply each numerator into all the denom- | 
inators except itsown, for a new numerator. Then mul- 
tiply all the denominators together for a new denomination | 
and place it under each new numerator. | 

RULE 2nd. Find the least common multiple of the given | 
denominators for the common denominator; then divide | 
the common denominator by each given denominator and 
multiply the quotient by its given numerator; the several 
products will be the several new numerators.- (See 
PROBLEM X, page 23.) 

The ist. of the above rules is convenient when m | 
terms of the fractions are small numbers, but the 2nd. is; 
otherwise to be preferred, as it always gives a denomina- 
tor which is the least possible. Other methods of finding 
a common denominator will occur to the student, after 
further practice. ‘4 

If any of the fractions to be reduced to a common de- 
nominator be compound, they must first be simplified. — 

37. Reduce 3, 44, =4 and 42 to a common denomi- 
nator. 

Tn this example, the least common denominator is found | 
to be 840. Then the several numerators of the common 
denominator are found as follows. 


| 
840+ 8—105, and 105 5==525. Ans. 2—228) 
340 + 12—= 70, and 7011770. =o 
840+ 14=—= 60, and 60 9=540. 38 
840+15= 56, ane 56 X 13= 728. ee | 


38. Reduce 3 7 +s and 3 to a common denominator. 
39. Reduce 2 oe 4354 and 5to a common denominator. | 
40. Reduce 24 3 eT and 42 2 to. a common denominator. 
41. Reduce 4 and 2 of + te a common denominator. 


t 
4 
| 
x. FRACTIONS. 48 


CASE IX. To reduce a complex fraction to a simple 
raction. 

RULE. Jf the numerator or denominator, or both, be 
vhole or mixed numbers, reduce them to improper frac- 
tons: multiply the denominator of the lower fraction into 
he numerator of the upper, for a new numerator; and 
nultiply the denominator of the upper fraction into the 
) aie of the lower, for a new denominator. 


42. Reduce 2. to a simple fraction. 
7 


| 9 
‘Th mation, os OS Ans. 88 
€ operation =5 1533 = 93 Ans 2 


48. geet a each of the following complex fractions. 


10 
53, 32, 21, 65, 28, 
Ale > ets g 


coins | 
° 


ADDITION OF FRACTIONS. 


Fractions are added by mere-y adding their numera- 
ors, but they must be of the same integers; we cannot 
nmediately add together 3 of a yard and 2 of an inch, 
sr the same reasons that we cannot immediately add 
gether 5 yards and 3 inches. They must, also, be of 
ie same denomination; we cannot immediately add to- 
ether fourths and fifths. 

RULE. Reduce compound fractions, (if there be any), 
» simple fractions, and reduce all to a common denomi- 
ator; then add together the numerators, and place their 
i um over the common denominator. If the result be an 

nproper fraction, reduce it to a whole or mixed number. 
1 44. Add together, 33, 3, 8$ and 4. 
| By operations not here de- 


I 

( 360 noted, we find the common de- 

ey “280 nominator to be 360; and also 

A 3 935 find the several new numerators. 

, 82 216 The sum of the fractions is 3¢5 

, 2 wi a [= — 2271, which, added to the 
Q271 9919271 | whole numbers; gives the total 

360 380: Re SEO} 


(sum, 13444 


44 ARITHMETIC. xX 


45. Add together, 97, 1244, 74, 2 and 213. 

46. What is the sum of -+2-+4$+2+4? 

47. What is the sum of 19,3,-+7 0 

48. What is the sum of 4 of 4 

49. Find the sum of $ of a sh 

In this example, first reduce the 3 0 

and the fraction of a penny. 

50. Find the sum of -?; of a gallon and Z ofa gill. | 

51. What is the sum of 52 days and 52,3, minutes ? | 

52. Whatis the sum of 4 of a cwt., 82Ib. and 3%, 02.| 
| 


@ of a penny ! 
a shilling to pence, 


—_— 


: 
SUBTRACTION OF FRACTIONS. 

As in addition of fractions we find the sum of the} 
numerators, so in subtraction of fractions we find th 
difference of their numerators. 

RULE. Jf either quantity be a compound fraction, ré 
duce it to a simple fraction, and if the two fractions hav’ 
different denominators, reduce them to a common denomh 
‘nator. Subtract the numerator of the subtrahend fror 
the numerator of the minuend, and place the remainde: 
over the common denominator. co ug 

When the minuend is a mized number, and the frac 
tion im the subtrahend is greater than that in the minu 
end, subtract the numerator of the subtrahend from th 
denominator, and to the difference add the numerator 0 
the minuend; and consider the integer of the minuen 
to be 1 less than it stands. 

It is not always obvious, which of two fractions ex) 
presses the greater quantity. In such case, the fraction 
are denoted with a character between them, thus, 53 DT 
and the greater is discovered by reducing them to acom| 
mon denominator. , a | 

53. What is the difference between 247 and 262? 


72 Here the fraction in the su 

3 ee trahend: is the greater, and w 

263 27 ne | 

947 56 are obliged to convert a unit int 
: ie seventy-seconds to obtain a quan 

143 43 by i 
ae Te tity from which to subtract 3$. 


54. What is the difference between ;2; from 18? 


xX, FRACTIONS. 45 


55. Perform subtraction on 33 wii. 

56. What will remain if 5145 be taken from 842? 

57. Subtract $ of 7 from 36-7. 

58. What is the difference between 44), and 10? 
' 59. What will remain if 2 of Z be taken from a unit? 
| 60. What is the difference between #7; and 13? 
61. 43— 5 of ¢ of 2 is equal to what quantity? 


MULTIPLICATION OF FRACTIONS. 


_ The following rules for multiplication of fractions, are 
vased on the Propositions 1, and 11, stated in page 37. 


, CASEI. To multiply a fraction by a whole number. 
| RULE. LHither multiply the numerator, or divide the 
enominator by the whole number. 


| CASE TI. To multiply a whole number by a fraction. 
) RULE. JSMultiply the whole number by the numerator, 
nd divide the product by the denominator. 


(CASE II. ‘To multiply a fraction by a fraction. 
| RULE. Multiply numerator by numerator, and denom- 
ator by denominator, for a new fraction. 


' When both factors are mixed numbers, it is generally 
‘lore convenient to reduce them to improper fractions 
‘nd then proceed according to the rule under Case itr. 
| The effect of multiplying any quantity by a proper 
‘action is, to give in the product, such a part of the 
uantity multiplied as the fraction indicates. Thus the 
‘roduct must be less than the multiplicand. _ This effect 
tthe operation will appear consistent with the principle of 
uultiplication, when it is considered, that multiplying any 
umber by 1, gives only the same number in the pro- 
uct; and, therefore, multiplying by less than 1, must give 

product less than the number multiplied. 

, 62. Multiply23by 9. 23x9 — 25X92 — 25 — 82 

| 63. Multiply 49 by &. (See rule under Case 11.) 

| 64. Multiply 35 by 2. (See rule under Case 111.) 
65. Multiply 6% by 344. (Remark under Rule m1.) 
66. What is the product of 34 by 15? 


AG ARITHMETIC. | 


| 
67. What is the product of 9241 by = fo! | 
68. What is the product of ¥ z, by 23? | 
69. What is the product of i 5 by 12%? RY 
70. Which is the most, 3X 65, or, 65 xo? a 
71. What is the product of 2941: 3 by 25? f 
In this example, it will be most convenient to find tl 
product of the whole numbers without regard to the fra 
tion first; then find the product of the fraction in a sep) 
rate operation, and, finally, add the two products togethe! 
72. What is the product of 361 by 3443? 
73. How many square inches of paper “in a sheet th 
is 142 inches long, and 113 mches wide? 


DIVISION OF FRACTIONS. | 


The rules for division of fractions, like those for a | 
plication, are ceey on Propositions 1, and 11. | 


| 
CASEI. To Taiie a fraction by a whole number. 

RULE. Hither divide the numerator, or multiply ti 
denominator, by the whole number. 


CASE II. To divide a whole number by a fraction. | 1 
RULE. Multiply the whole number by the denomim) 
tor, and divide the product by the numerator. | 


CASE 111. To divide a fraction by a fraction. 

RULE. Invert the divisor, and then proceed as in mu 
tiplying a fraction by a fraction. | 

Observe, that the operation of this last rule is, to mu} 
tiply the denominator of the dividend by the numeraty, 
of the divisor for a new denominator, and the numerat| 
of the dividend by the denominator of the divisor | 
new numerator. 

Compound fractions are to be reduced to simple one, 
and mixed numbers to improper fractions, before th 
adoption of either of the above rules. | 


74. Divide 7 by 8. 7+S—=asieg=—ay. Ans. & 
75. Divide 14 by +. (See rule under Case 11.) © 


76. Divide 2; by 2 ‘(See rule under Case III.) 
77. Divide the compound fraction 3 2 of 2, by 6. 


j 


XK. FRACTIONS. AT 


78. Divide 325 by the mixed number 53. 
79. What is the quotient of $4 divided by 13? — 
80. What is the quotient of 57 divided by ;5? 
81. What is the quotient of 4 divided by 4? 


82. Divide § of 5 by % of $ of 2. 
| 83. What is the quotient of 912 divided by 15? 


| 84. What is the quotient of 2062 divided by 944? 

. 85. How many times is 24 contained in 319? 

- 86. How many times is 194 contained in 994? 

87. How many times ¢ of an inch in =; of a yard? 
First, reduce the 3°, of a yd. to the fraction of an inch. 
88. How many times 2 of a gill in 3 barrels ? 

89. Suppose a wheel to be 11,7; feet in circumference; 
1ow many times will it roll round in going 393 rods ? 


MISCELLANEOUS: EXAMPLES. 


In the following examples, all fractions which appear 
athe answers, must be reduced to their value in whole 
tumbers of lower denominations, whenever there is op- 
iortunity for such reduction. 

90. What distance will a carrun in 93 hours, allowing 
ts velocity to be 232 miles an hour? 

91. Suppose a car wheel to be 8 feet 7 inches in cir- 
uumference, how many times will it turn round in running 
165 miles ° 

92. If 32 cwt. of sugar be taken from a hogshead con- 
ainmg 14cwt. 1qt. 64lb., how much will remain in the 
iogshead ? - 

93. What is the sum of 16Zcwt., 7ewt. 3qr. 84ib., 
IT. 194cwt., 2cwt. liqr., and 7 of a ton? 

94. A farmer owning 1322 acres of land, sold 464A. 
3R.12r. How much land had he remaining ? 

_ 95. What is the value of 367 acres of land, at $472 
yer acre? 

| 96. What is the value of 154 barrels of flour, at 
$4.625 per barrel ? 

97. What is the value of a- load of wood, containing 
i feet, [§ of a cord,] at $5.25 per cord? Or, what is 3 
£ $5.25? Or, $5.25xXS—? 

) 


48 ARITHMETIC. 4 


3544 rods on every side? (See page 28.) 
99. What quantity of land in a lot, which is 654 ro 
long and 47} rods wide ? | 
100. What quantity of wood is there in a pile, 14- 
feet long, 342; feet wide, and 6,3; feet high: ? i 
101. Supiobs a lot of land to be 61rods wide, ho 
long must it be, to contain 1 acre? (See Pros. v, pa 


) 
98. How much land is there in a square lot, | 
. 
i 


21. Consider that 1 acre contains 160 rods.) 
102. What quantity of loaf sugar must be sold at 19 ; 
cents per pound, that the price shall amount to $524 | 

103. What cubical quantity of earth must be ae 
in digging a pit, 135 feet deep, 123 feet long, and § 
feet wide ? a) 

104. What quantity of hewn timber is there in a sti 
that is 124 feet long, 24 feet deep, and 12 foos wide ?) 

105. Suppose a stick of timber to be 14, 5 foot dee) 
and 8 inches wide; what must be the leneth oF the stic 
in order that its quantity shall be 1 ton of hewn tim) 
| (See Pros. viii, page 22. Consider a ton as the pr’ 
duct of three factors. ) 

106. Suppose wood to be piled ona base 18 feet lo 
and 73 feet wide, what must be the height of the pile, 
contain 91 cords? | 
107. What quantity of molasses in 4 casks, seal? 


severally, 55;gal., 31ljgal., 277; gal., and 584%, gal.? 
108. What i is the cost of 486 bushels of corn, at 69) 
cents per bushel ? 


ne 


109. Suppose 63. gallons to have leaked from a hog 
wine, at 874 cents per gallon? | 
110. How many bottles, each holding 13 pint, are rm 
111. Suppose 44 gallons of cider to have evaporate 
from a barrel; what number of bottles, each holding 1p 
112. What is the value of 1425 tons of coal, at 7 
dollars per ton? 
the rate of 2 of a dollar per bushel? _ Xf?) a 
é eh. ! 


head of wine, what 4 is the value of the remainder of a 
quired for bottling 3 barrels of cider? | 
3sel., will be required to bottle the remainder ? 
113. What is the value of % of abushel of wheat, i) 
: Ks eae, | 
| 


x. FRACTIONS. 49 


ie 114. If 1 hogshead [63 gal.| of molasses cost $262, 
what is the cost of 1 gallon ? 
/ 115. What is the cost of 7hhd. 63 gal. molasses, at 
1, cents per gallon? 
_ 116. What is the cost of 25 yards 34 2 quarters of rib- 
bon, at 194 cents per yard? 
mys If 54 + cords of wood cost $265 , what is the cost 
lof 1 cord? 
118. What is the value of 162 tons of hay, at 114 
dollars per ton? 

119. What is the value of 1lb. 60z. 12 dwt. of silver, 
te 20: cents per pennyweight ? 
| 120. If 167 yards of broad-cloth cost $86.24, what 
is the cost of 1 yard? 
» 121. At 5s. 34d. per yard, what is the cost of 783 
yards of cambric, in pounds, shillings, and pence? 
| 122. If 49232 yards oftcloth cost £68 4s. 10d., what 
jis the cost of 1 yard? 
/ 123. If 183-yards of cotton cost 12s. 9d., what is the 
‘cost of 1 yard? 

124. What is the value of 57683]b. of coffee at 102 
;pence per pound ? 
| 125. At what price per pound must I sell 4323 pounds 
of coffee, in order to receive £27 3s. for the w hole: P 
| 126. If £448,5 be equally divided among 76 ai 
what will each man receive ? 
127. If ofa yard ~ eee eost $3, what is the price 
of 1 yard? on Go 
fe 198. If 743 barrels i stow cost $214, what is the 
‘cost of 1 barrel of the apples ? 

129. If 44 gallons of molasses cost $23, what is the 
‘cost of 1 quart: P 

130. If 14 hogshead of wine cost $2504 what is the 
icost of 1 quart : ° 
| -131. Bought 5 yards of sine at $24 per yard; 154 
yards of ribbon, at 124 cents per yard; 17 pairs of gloves, 
at 68% cents per pair; ‘and 164 1 yards of lace, at $34 per 
yard. What is the whole cost ? 
| 132. Bought 64 pounds of tea, at 875 cents per pound; 
153 pounds of su oe 4 113 cents per pound; 13% pounds 


= 


50 ARITHMETIC. x 


of coffee at 125 cents per pound; and 163 gallons ol 
molasses, at 4 of a dollar per gallon. What is the whok 
cost ? 

133. Bought 93 barrels of cider, at $24 per barrel 

8 barrels of apples, at $12 per barrel; 16 boxes of raisins. 
at $2.624 per box; 234 pounds of almonds, at 142 cent: 
per pound. What is the whole cost ? 
_ 134. Bought 3584 bushels of wheat, at $ ofa dolla] 
per bushel; 420 bushels of rye, at 964 cents per bushel 
1464 bushels of corn, at 2 of a dollar per bushel; ane} 
6511 bushels of oats, at 232 cents per bushel. What is 
the whole cost ? 

135. A purchased of B, 752 tons of iron at $9. 614 
perton. What quantity of coffee, at 123 cents per pound, 
must A sell B, to cancel the price of the iron? 

136. © purchased of D, 1397 hogsheads of molasses, 
at 153 cents per gallon; and D, at the same time, pur. 
chased: of C, 896) tons of iron, at $94 per ton. How 
much was the balance— and to raheniy was it due? 

137. What is the sum of 4 13> qa %, 34,45, 42, 7 
$9 50 99 479 tp eodigy of et | 

138. Suppose +; of 3% of 74 to be a minuend, and 
of 2-of t of # a subtrahend; what is the remainder? 

139. What is the product of 4 of 2 of ¢ of 100, multi 
plied by 4 of 2 of Z of 20f 752 - 

140, What is the quotient of 2 of 2 of 32, divided by! 
+ of 42 of 2 of 32 of 2? 

141. Suppose the sum % two fractions to be 2, and’ 
one of the fractions to be 353; what is the other? (See| 
PROBLEM I, page 20.) 

142. Suppose the greater of two fractions to be 44,) 
and their difference to be 4 43 ; what is the smaller fraction? 
(See Pros. 11, page 20.) ( 

143. Suppose the smaller of two fractions to be 34, 
and their difference to be 3; what is the greater fraction | 
(See Pros. 111, page 21.) 

144. What are the two fractions, whose sum is 42, ari 
whose difference.1 is f7? (See Pros. iv, page 21.) | 

145. If .°5 be the product of two factors, one of which’ 
is 77, What is the other? (See Pros. v, page 21.) 


SS 


—— 


en haa = 


XI. DECIMALS. 51 
» 146. Suppaseny % to be a dividend, and <5 a quotient; 
what is the divisor? (See Pros. vi, page 21.) 

147. What must be that div iead: whose divisor is 
22 and whose quotient is 3? (See Pros. vii, page 22. 3) 
, 148. Suppose the product of three factors to be 32, 
one of those factors being 2, and another 7; ; what is the 

tae third factor? (See Pons VIII, page 22. iP 
149. A merchant owning 3% of a ‘ship, sold 2 of what 
he owned. What part of the whole ship did he sell ? 
} 150. A merchant owning 32 of a ship, sold $ of what 
he owned. What kee of the ship did he still own? 
| 151. If I buy = of 3 of a ship, and sell 2 of what I 
‘bought, what part ‘of the ship shall I have left? 


| The kind of fractions, which have been treated in this 
article, are called Vulgar fractions, or Common frac- 
“tons, in distinction from another kind, called Decimal 
fractions, or simply Decimals. 


* 


dF 
DECIMAL FRACTIONS. 


A DECIMAL FRACTION is a fraction whose denominator 
jis 10, or 100, or 1000, &c. The denominator of a decimal 
fraction is never written: the numerator is written with a 
point prefixed to it, and the denominator is understood 
ito be a unit, with as many ciphers Rpu ered as the le 
ator has places of figures. Thus, .5 is 7%, .26 is #5, 
907 is #95 


1000 ° 
When a whole number and decimal fraction are written 


together, the decimal point is placed between them. 
Thus, 68.2 is 6875, 4.87 is 4fr. 

In the notation of whole es any figure, wherever 
3 may stand, expresses a quantity 7/5 as great as it would 
express if it were written one place further to the left: 
and so it isin the notation of decimal fractions—the same 
system is continued below the place of units. The first 
place to the right of units is the place of tenths; the second, 


52 ARITHMETIC. XI 


of hundredths; the third, of thousandths; the fourth, : 
ten-thousandths; and so on. 

Ciphers placed on the right hand of decimal figures, és 
not alter the value of the decimal; because, the figures 
still remain unchanged in their distariee from the unit’s! 
place. For imstance, .5,°.50, and .500 are all of equal 
value,— they are each equal to 5. But every cipher that 
is placed on the left of a decimal, renders its value ten 
times smaller, by removing the ficures one place further 
ats the unit’s place. Thus, if we prefix one cipher to 

5, it becomes .05 [725]; if we prefix two ciphers, it 
becomes .005 [550]; and so on. ‘ 


To READ DECIMAL FRACTIONS—Enumerate and read 
the figures as they would be read if they were whole num= 
bers, and conclude by pronouncing the name of the lowest 
denominalton. ‘| 


1. Read the several numbers in the following columns. 


| 
| 
.99 .2008 4.008 24.09 | 
.064 -00006 §.37002 630.1174 } 
0003 ~=—s-«.03795~=——S~St«é«C 9999 6.972479 | 
5237 .180009_-. 5.0001 28.797 
2. Write in decimals the following mixed numbers. 
Sehey 21560 «© 8800 Sisco 
24790 326 750 Siot00 © 47 ro0000 
38 7000 T9000 «= 97 zdd0 6 x5000 
CSaps00— sW9TaSGa CO too | OS tan 


10000000 | 


ADDITION OF DECIMALS. 


3. Add the following numbers into one sum. 1051. 7 
+ 70.602 -++ 4.06 + 807.2659. 


151.7 — In arranging decimals for addition, | 
70.602 | weplace tenths under tenths, hun- 
4.06 dredths under hundredths, &c. We 


_ 807.2659 then begin with the lowest denomi-| 
"1033.697 9 nation, and proceed to add the col- 
say yumnns as in whole numbers. { 


i 


; 
) 


XI. DECIMALS. 53 


4. What is the sum of 256.94-+-9121.7-+- 8.3065? 
5. Add together .6517-+ 19.2-+ 2.8009 + 51.0007 + 
00009 + 22.206 +-4.732. 
In Federal Money, the dollar is the unit; that is, dol- 
lars are whole numbers; dimes are tenths, cents are 
hundredths, and mills are thousandths. 
‘2 6. Add together $18.25, $4.09, $2.40, $231.075, 
,$64.207, $50.258, $10. 09 and ets. 
\ ae Write the following sums of money in the form of 
decimals, and add them together. $1 and Icent, 37 
cents, $25 and 7 chimes, 65 cents, $15, 9 dimes, 8 nulls; 
4 cents and 3 mills, -§ of a mill, 7 and 8 cents, 3° of a 
mill, 36; cents, 10 eagles and 25 dollars, and 7 cents. 
} 


SUBTRACTION OF DECIMALS. 
_ 8. Subtract 4.16482 from 19.375. 


19.375 After placing tenths under tenths, 
4.16482 &c., we subtract as in whole num- 
15.21018 bers. ‘The blank places over the 2 


ee + and (S are: viewed<asi ciphers; 

9. Subtract 592.64 from 617.23169. 

10. Subtract 48.06 from 260.3. 

11. Subtract .89275 from 12690.2. 

12. Subtract .281036 from 51. 

13. What is the difference between 1 and .1? 

14. What is the difference between 24.367 and 13? 

15. What is the difference between .136 and .1295? 
_ 16. Write 8 dollars and 7 cents in decimal form, and 
‘subtract therefrom, 48 cents and 1 mill. 

17. Subtract 9 dimes and 6 mills from 15 dollars. 


MULTIPLICATION OF DECIMALS. 


Multiplying by any fraction, is taking a certain part of 
the multiplicand for the product; consequently, multiply- 
‘ing one fraction by another, must produce. a Faction 
| smaller than either of the factors. For example, +475 
|\==005 Or, decimally, .4x.3—.12. Hence observe, 

that the number of decimal figures in any product, must 
| 5x 


7 ARITHMETIC. XL 


be equal to the number of decimal figures in both the) 
factors of that product. t 

RULE. Multiply as in whole numbers; and im the 
product, point off as many figures for decimals, as there 
are decimal places in both factors. Tf the number of 

gures in the product be less than the number of decimal 
places in both factors, prefix ciphers to supply the si) | 
ciency. 


18. Find the product of 658 by .249. 7.06 by 3.05. 
593 by 5.62. .146 by .244. 


————$—$$—. — +. ———— oo es 


658 7.06 5.93 146 
249 3.65 562 +244 
5922 3530 1186 584 
2632 4236 3558 584 
bgl6e 2118 2965 292 
173.842 25.7690 3.33266 035624 


19. Multiply 428 by .27; that is, find .27 of 428. 
_ 20. What is the product of 3.067 by 8.2? 

21. What is the product of .6247 by 23? 

22. What is the product of .099 by .04? E | 

23. What is the product of .113 by .0647 ? | 

24. What is 7.03 X .9 x 31.6 XK 28.758 =? | 

25. Multiply 9dolls. 7 cts. 6mills [9.076] by 46. — 

26. What cost 28 yards of cloth, at $'7.515 per yd.? | 

27. What cost 15.9 yd. of cloth, at $9.427 per yd.? 

28. What cost 275 lemons, at 9 ‘mills apiece f a 

29. At 7 cents and 3 mills per yard, what is the value) 
of 18704 yards of satin ribbon ? e 

30. What is the value of a township containing 30519.75 
acres of land, at 4dolls. 8cts. and 5 mills per acre? 

31. What is .06 of 1532 dollars? Or, what is the) 
product of 1532 multiplied by .06? | 

32. What is 03 of 476 ashi and 78 cents ? S| 

33. If an insurance office charge .015 of the value of a. 
house for insuring it against fire, what will be the exper 
of insuring a house, valued at $437.25 ? 

34. Multiply 26.000375 by .00007. 

35. What is the product of 3.62981 by 10000. 


~ 


- 
‘ 


XL. DECIMALS. 55 


_ The learner will perceive, that any decimal number is 
nultiplied by 10, 100, 1000, &c., by merely removing 
the decimal point as many places to the right hand as there 
awe ciphers in the multiplier. Thus, 6.25 x 1062.5. 
5.25 x 1000 — 6250. 

| DIVISION OF DECIMALS. 


_ It has been shown, in multiplication of decimals, that 
here must be as many decimal places in a product as 
here are in both its factors; and it follows, that, in divi- 
sion of decimals, there must be as many decimal places 
nthe divisor and quotient together, as there are in the 
dividend. Therefore, the number of decimal places in 
he quotient must be equal to the difference between the 
vumber of decimal places in the dividend, and the num- 
yer of decimal places in the divisor. 

RULE Divide as in whole numbers; and in the quo- 
lent, point off as many figures for decimals, as the deci- 
nal places in the dividend exceed those in the divisor; 
hat is, make the decimal places in the divisor and quotient 
sounted together, equal to the decimal phones, in the divi- 
lend. 

If there be not figures enough in the Hatta to notnt 

Off, prefix ciphers to supply the deficiency. 

When there are more decimal places in the divisor, 
han in the dividend, render the places equal, by annez- 
mg ciphers to the dwidend, before dividing. 

After dividing all the figures in the dividend, if there 
ve a remainder, ciphers may be annexed to ti, and the 
lwision continued. The ciphers thus annexed, must be 
sounied with the decimal places of the dividend. 


36. How many times is 57.2 contained in 2406.976 ? 
57.2)2406.976(42.08 
| 37. What is the quotient of 11.7348 by 254° 
| 254)11.7348(.0462 
| 38. What is the quotient of 4066.2 by .648? 
| 648)4066.200 (6275 


é 


56 ARITHMETIC. 7 


39. What is the quotient of 3.672 by .81° 
| The sign of addition, 
-81)3.672(4 5333-+- | more, here shows, that the trur 


324 quotient is more than the pre, 
432 ceding figures express. We 
405 .. Sigeught continue the division 
O70 but we should never arrive a 

943 a complete quotient. For thc 
ee purposes of business, it is sel’ 
270 dom necessary to extend the 
243 quotient below thousandths 
270 but, in the following exercises, 
243 those quotients that do no 
oF terminate, may be extended te 


illGuths! | 
40. How many times is 4.72 contained in 637.531 4 
41. What is the quotient of 2.7315 by 74° | 
42, What is the quotient of 409.867 by .5806° 
43. What is the quotient of 125 by .1045? . 
44, What is the quotient of 709 by 3.574? | 
45. What is the quotient of 7382.54 by 6.4252? | | 
46. What is the quotient of 715 by .3075 ? ; 
47. What is the quotient of 267.15975 by 13.25? —| 
48. What is the quotient of .0851648 by 423? | 
49. What is the quotient of .009 by .00016? bY 
50. If 17 boxes of oranges cost $98.29, what is the 
cost of a single box ? | 
51. If $550.725 be divided equally among 15 men) 
what will be each man’s share ? 
52. If 37.5 barrels of flour be divided equally anon 
25 men, how much will each man have ? +| 
53. If 46.75 yards of cloth cost $251. 702, what j i) 
the cost of 1 yard of the cloth? 
54. Divide 3712 by 42; annexing ciphers to the re 
mainders, until eight decimal figures are obtained in the 
si cpges | 
What is the quotient of 9 divided by 266? 4 


ms this example it will be necessary to annex a sufli 
cient number of decimal ciphers to the dividend, belgie 
the operation of dividing can be commenced, 


XI. DECIMALS. 57 


56. What is the quotient 1 divided by 8? 
57. What is the quotient of 62 divided by 97? ' 
' 58. Divide 1 by 2, 3by 4. 10 by 12. 3 by 16. 
Yby13. 6 by26. 14 by15. 40 by72. 7 by 599. 

_ Any decimal number is divided by 10, 100, 1000, &c. 
vy merely removing the decimal point as many places to 
he left hand as there are ciphers in the divisor. Thus 
4.8+10=1.48 14.8+1000—.0148 

| REDUCTION OF DECIMALS. 

CASEI. To reduce a vulgar fraction to a decimal. 
RULE. Divide the numerator by the denominator, and 
he quotient will be the decimal. 

59. Reduce { to a decimal. 

8)7.000 Decimal ciphers are here annexed 
to the dividend as directed in the 
875 Ans. ‘rule for division of decimals. 

60. Reduce the fractions 4, 2, 4, 7, +4, 4, 3) 
nd +755 to decimals. 

61. Simplify 2 of =4, and reduce it to a decimal. 

62. Reduce 1 of 2 of 4 to a decimal. 

63. What is the decimal expression of 24754? 

64. Reduce 3, 37, and 5, to decimals. 

The learner will discover, that the above fractions, 2, 
2, and = cannot be reduced to exact decimal expres- 
ions. The quotient of 2 by 3 is .6666, &c., continually. 
“he quotient of 2 by 11 is .181818, &c.; the same two 
gures being repeated continually. ‘The quotient of 1 by 
'7As .037037, &c.; the same three figures being repeated 
ontinually. Decimals of this kind are treated in the 
ext Article, under the head of Infinite Decimals. For 
10st purposes, however, three or four'decimal places will 
xpress any fraction with sufficient accuracy, unless the 
iteger of the fraction is of very high value. 


| CASE II. To reduce a decimal to a vulgar fraction. 
RULE. Write the decimal denominator under the dec- 
mal, and erase the decimal point: view the expression 
_savulgar fraction, and reduce it to us lowest terms 


ms 
ee 
me HES 


58 ARIVHMETIC. XI 


65. Reduce .4375 to a vulgar fraction. | 
.4375 — 7434, ; and to reduce this fraction to its lowes 
terms, we divide the terms by their greatest commo, 
measure, which is 265. The result is, 74. 41, 
66. Reduce .375 to a vulgar fraction. 5 | 
67. Reduce .76482 to a vulgar fraction. I 
68. Reduce .510505 to a vulgar fraction. 4 
69. Reduce .1084058 to a vulgar fraction. | 
70. Reduce .04608128 to a vulgar fraction. 


CASE 10. ‘To reduce the lower denominations of 
compound number to the decimal of a higher denomina, 
tion. 

RULE. Reduce the given quantity toa vulgar fraction 
(as taught in page 40), then reduce the vulgar fraction, 
to a decimal. 

The decimal quotients which do not terminate, may) 
in the examples of this case, be extended as low as thi 
seventh place. 

71. Reduce 12s. 6d. 3qr. to the decimal ofa £. | | 

72. Reduce 2qr. 14|b. to the decimal of a cwt. 

73. Reduce 1R. 14 rods to the decimal of an acre. | 

74. Reduce 13 dwt. 16 gr. to the decimal of a pound. 
Troy weight. i 

75. Reduce I pk. 1 pt. to the decimal of a bushel. | 

76. Reduce 1bl. to the decimal of a ton of wine. | 

77 Reduce 4yd. 6in. to the decimal ofa mile. 

78. Reduce 5 square yards to the decimal of an acre’ 

79. Reduce 14 cubic feet to the decimal of a cord. | 

80. Reduce 21h. 50m. 31s. to the decimal of a year’ 

81. Express £19. 13s. 93d. decimally; making | 
£ the unit, and the s. and d. a decimal. | 

82. Reduce 17hhd. 9 gal. 3qt. 1 pt. to a decimal ex 
pression; the hogshead being the unit. 

83. Reduce 15 tons, 1qr. 1402. to a decimal expr 
sion; the ton being the unit. 

84. Reduce 4 miles, 7fur. 9r. 3yd. Gin. to a decima, 
expression; the mile being the unit. i. | 

85. Reduce 25 rods,19yd.7 ft. 115in., square measure; 
to a decimal expression; the rod being the unit. | 
| 
| 


xT. DECIMALS. 59 


86. What is the value of feist) ahs 1 pt. of wine, at 
eh 359 per gallon? 

In this example, first ae the quantity of wine toa 
Yecimal expression ,— the gallon being the unit— and then 
nultiply this quantity into the price of 1 gallon: the answer 
will be $566.533-+-. The following-examples in this 
sase are to be performed in like manner. 

87. What is the value of 57yd. 2qr. 3 na. of cloth, at 
$6.78 per yard? 

88. What is the value of 7483 yards of ribbon, at 9 
‘ents 8 mills [.098] per yard? 
| 89. What is the value of 5741 yd. Sqr. of tape, at 7 
nills [.007] per yard? 

: 90. What is the value of 4cwt. Lqr. 19|b. of raisins, 
it $12 per hundred-weight ? . 

91. What is the value of 32 hhd. 22gal. of molasses, 
t $19.22 per hogshead ? 
| 92. What is the value of 3pk. 7 qt. of corn, at'75 cents 
ver bushel ? 

93. What is the cost of 15E.e. 4qr. 3na. of linen, 
t $1.15 per ell? 

94. What is the cost of 7A. 2R. 38r. of land, at 
.64.50 per acre? 

95. What is the cost of 28 square rods and 260 square 
»et of land, at $84.25 per rod ? 

96. What is the cost of 29lb. 60z. 8dr. of indigo, at 
}3.75 per pound ? 
| 97. What is the cost of 4qr. 3na. of thread lace, at 
4. 50 per French ell.? 

| 98. What is the value of 7lb. 100z. 18 dwt. of copper, 
| 27 cents per pound ? 

99. What is the value of 11 oz. 19 dwt. 23 gr. of silver. 
t $15.25 per pound ? 
| CASE Iv. To reduce the decimal of a higher denomi- 
ation to its value in whole numbers of lower denomina- 
on. 

RULE. Multiply the tecthita) by that number of the 
ext lower denomination which makes a unit of the high- 
r, and the product will be of the lower denomination. 


a 


j 
60 ARITHMETIC. XI 


Proceed in like manner with the decimal in each sue 
ceeding product. — ‘| 


100. Reduce .769r. to its value in yards, feet, ani 
inches ; ; that is, change .769 of a rod to yards, &c. 
769 There are 54 2 times, or 5.5 time, 
ars as many yards as  rodsin any quantity: 
whether that quantity be a whole num| 
3845 ber or a decimal: therefore, we mul: 
3845 tiply the decimal of arod by 5.5, ani 
4.2295 the product is 4.2295 yards. Wi 
3 then multiply .3295 of a yard by 35t) 
6885 find the feet; but there is not a whol 
12 foot in this decimal, and we procee 
E to find the inches. The whole resu. 

8.2620 is, 4yd. Oft. 8.262in. 


101. Reduce .775 £ to its value in shillings, &c. | 
102. Reduce .625s. to its value in pence and farthings 
103. Reduce .46941b. Troy, to oz., dwt., &c. 

104. Reduce .624cwt. to its value in qr., ‘lb., &ec. 

105. Reduce .0653 mile to its value in yd, &e. 

106. Reduce .3875 A. to its value in R. andl rods. 1 

107. Reduce .0098 ton to its value in lb., oz., and di) 

108. Reduce .2083hhd. to its value in gallons. 4 

109. Reduce .467 cwt. to its value in qr. lb. &c. ¢ 

110. Reduce £741.687 to its proper expression, J 
pounds, shillings, pence, and farthings. 

111. Reduce 84.704 miles to its proper expression 
in miles, furlongs, rods, yards, &c. 

112. Reduce 50.742A. to its proper expression, } 
the several denominations of square measure. 4) 
| 
| 


i 
¥ 


EXCHANGE OF CURRENCIES. 


Before the adoption of the Federal currency, merch 
in this country, kept their accounts in the denominatior 
of English money. The value of the Pound, howevei 
and consequently the value of its subdivisions, was val) 
ous: that is, a pound, and consequently a shilling, signifie! 
a greater value of money in some of the states, than j i 
others. Accounts are now kept, in Federal money, an 


XI. DECIMALS. — 61 
its denominations are generally used in stating prices. In 
some sections of the country, however, prices are fre- 
quently mentioned in shillings and pence—a custom which 
is inconvenient, and which ought to be discontinued. 
» In New England, Virginia, Kentucky, and Tennes- ~ 
see, ¢ of a dollar is called a shilling. 

In New York and North Carolina, } of a dollar is 
called a shilling. . oe 
' In Pennsylvania, New Jersey, Delaware, and Mary- 
land, 7; of a dollar is called a shilling. 
| In South Carolina and Georgia, 3%; of a dollar is call- 
ied a shilling. 
' “In Canada, + of a dollar is called a shilling. 


113. How many cents and mills, that is, what decimal 
of a dollar, ina New-England shilling? in 2 shillings ? 
in 8 shillings? in 4 shillings? in 5 shillings ? 

114. How many cents and mills in a New-York shil- 
Img? in 2s.? in 3s.? in4s.? m5s.? in 6s.? in7s.?° 

115. How many cents and mills ina Pennsylvania shil- 
ling? in2s.? in 3s.? in4s.? in 5s.? in 68.? 
| 116. How many cents and mills in a Georgia shilling ° 
m2s.?:in3s.? in4s.° 

17. How many cents are there in a Canada shilling ? 
fme2s.° in 3s.? m4s.p in 5si? 
| To change the old currencies to Federal money. 

RULE. Reduce the pounds, if there be any, to shil- 
\lings. Denote the shillings as wnits, reduce the pence 
and farthings to the decimal of a shilling, and multiply 
the whole swm by that fraction of a dollar which ts equal 
to one shilling. n 

118. Change 13s. 6d., of the old currency of New 
| England, to Federal money. 
| 119. Change £42 19s. 43d. of the old currency of 
b England, to Federal money. * 

120. Change 13s. 6d., of the old currency of New 
| York, to Federal money. 
| 121. Change £25 17s. 83d., of the old currency of 
| New York, to Federal money. 

| 6 


| 


62 ARITHMETIC. XI. 


122. Change 18s. 11d., of the old currency of Penn- 
sylvania, to Federal money, 

123. Change £14 7s. 64d. of the old currency of 
Pennsylvania, to Federal money. 

124. Change 16s. 10d., of the old currency of Georgia, 
to Federal money. | 

125. Change £54 12s. 114d., of the old a 
of Georgia, to Federal money. | 

126. Change 17s. 5d., of the currency of Canada, to 
Federal money. . 

127. Change £21 9s. 33d., of the currency of 
Canada, to Federal money. 

128. What is the value, in Federal mpons of 9 New 
England shillings? 9 New York shillings? 9 Pennsyl- 
yania shillings? 9 Georgia shillings? 9 Canada shillings? 


J 
2 


MISCELLANEOUS EXAMPLES. 


Any vulgar fraction, which shall appear in the following 
examples, must be reduced to a decimal; and the lower) 
denominations of compound numbers must also be reduced: 
to decimals, before they are brought into operation. No: 
decimal need be continued lower than siz places. Anz! 
swers to be given in decimals. 7 

129. What is the sum of 6 tons 18cwt. Lqr., 5 owt) 
Sqr. 2lb., 4.093825 tons, 2qr. 27]b., Scwt. Qqr. 4 lb. S| 

and 17 tons 5 cwt. Oqr. 19]lb.? 

130. What is the difference between 2. 90843hhd. | 
and 4hhd. 47 gal. 3qt. Ipt. of wine? 

131. What is the cost of 15.179 yards of broadclotfil 
at $6 per yard? 

132. If 57 yards of cloth cost $197, what costs 1 yd.?' 

133. What is the cost of 28 yd. 3qr. “of cloth, at ¢ 7. 55 

er yard? 

134. If-18yd. 1qr. of cloth cost $91.16, what is the 
cost of 1 yard? 

135. What is the cost of 25cwt. 2qr. 20Ib. of hops,’ 
at $4.96 per hundred weight ? | 

136. What is the cost of 24hhd. 15gal. of molasses, 
at $25.36 per hogshead ? | 


XI. DECIMALS. 63 


_ 137. What is the cost of 256 yd. 3qr. of ribbon, at 
8 cents 5 mills [.085] per yard? 

138. What is the cost of 244 yards of ribbon, at 7 
cents per yard? 
_ 139. What is the cost of 3qr. 2na. of broadcloth, at 
$10.35 per yard ? 
| 140. What is the cost pi lfir. 7 gal. 3qt. of beer, at 
$3.50 per firkin ? 
, 141. What is the value of 23 grains of silver, at $14 
per pound, Troy ? 
| 142. What is the value of 25 square rods of land, at 
$75 per acre? 
| 1438. If $238.86 be divided equally among 18 men, 
what will each man receive ? 

144. If $775 be divided equally among 8 men, what 
will each man receive ? 

145. If a man travel 73.487 miles in 15 hours, what 
distance does he travel in 1 hour? 
| 146. What is } of 1142.26? 
| The result will be the same, whether we divide 1142. 
26 by the denominator 8, (which is multiplying by §), 
or, reduce } to a decimal and multiply this decimal into 
1142.26. ‘The former method is to be preferred; and. 
the learner is here reminded, that the product of any 
decimal will be such a fractional part of the multiplicand 
as the decimal indicates. 

147. What is Ari of 1142.26? 

148. What is 4 of 2.565? (Divide by 6). 

149. What is .6 of 2.565? (Multiply by .6). 

150. What is + of 1999.2? 

151. What is .56 of 1999.2? 
/ 152. What is 745 of 387.65? 
| 7153. What is .135 of 387.65? 

154. What is 78, of 37241 dollars ? 
| 155. What is .06 of 37241 dollars ? 

156. Suppose I have $5872, and pay away .06 of it; 
how much shall I have left ? 

157. A owes B $430.40 to be paid in 10 months; 
but B relinquishes .05 of the debt for having it paid im- 
mediately. How much does B relinquish : 


. 


64 ARITHMETIC. Xi. 


158. C borrowed of D, $72.85, agreeing to pay it in 
16 months, increased by .08 of itself. What was the 
amount to be paid? 

159. What will it cost to insure a house, worth $ 2500, 
against the danger of fire, for one year, the price of 1 in- 
surance being .025 of the value of the house ? 

160. Suppose I purchase a ship for ¢$ 12900, and sell 
it at an advance equal to .019 of the cost; for how much 
do I sell it ? 

161.. How many gallons of wine can be purchased for 
$74, at $1.37 per gallon? 

162. How many pounds of raisms can be bought for 
$9, at 164 cents per pound ? 

163. if a man travel 5.385 miles in 1 hour, in how 
many hours will he travel 166 miles ? 

164. If 18 bushels 3 pecks of wheat sips on 1 acre, 
how many acres will produce 396 bushels ? 

165. If 3 shillings will pay for 1 bushel of barley, how 
many bushels will 26 shillings pay for ? 

166. If 5s. Sd. will pay for 1 bushel of wheat, how 
many bushels will £11 pay for? 

167. If 8s. 3d. will pay for 1 gallon of ‘ine how 
many gallons will £18 pay for ? 

168. What is the value, in Federal money, of £3 17 s. 
8d., of the old currency of New England? 

169. if i buy 230 pelts, in Canada, at 4s. 3d. api 
for what amount Federal money must I sell the wholly 
_ in the United States, in order to gain $36.15? 

170. How many square feet in a floor, that is 18. 8 
feet long, and 14ft. Sin. wide? 

171. How many square feet in a board, that is 16 ft | 
5in. long, and 11 inches wide?” . 

172. How many cubic feet in a box, that is 4ft. éin. 
long, Sft. 2in. deep, and 2ft. 9in. wide 2_ 

173. Goliath is said to have been 64 cubits high, adi 
cubit being 1 foot 7.168 inches. What was his height 
in feet? 

174. How many square feet of paper will it take to 0) 
cover the walls of a room, that is 18 ft. 9in. gad 14 ft. ft. 
6 in. re and 9ft. 3in. high? | 

| 


| 
| 


/ 


ae. DECIMALS. 65 


175 Suppose a man’s property to be worth $6520, 
and his tax to be .02 of the value of his property; how 
much is his tax ? 

176. If a man earn one dollar and one mill per day, 
how much will he earn in a year? 

_ 177 What is the cost of three hundred seventy-five 
‘thousandths of a cord of wood, at four dollars per cord? 
| 178. A has nine hundred thirty- six dollars, and B_ has 
five dollars, three dimes and one mill. How much more 
‘money has A than B? 

_ 179. A trader sold 4 pieces of cloth—the first con- 
tained 86 and 3-thousandths yards; the second, 47 and 3- 
tenths yards; the third, 91 and 7-hundredths yards; the 
fourth, 22 and 9-ten-thousandths yards. What did the 
whole amount to, at $7 per yard? 

180. A has $31.32, B has $572, C has $1042, and 
D has $954; and they agree to share their money equally. 
What must each relinquish, or receive ? 

181. Suppose a car wheel to be 2 feet 93 inches in 
circumference; how many rods will it run, in turning 
round 800 times ? 

182. If a car run 1 mile in 3 minutes and 9 seconds, 
in what time will it run 18 miles? 

183. Suppose the sum of two certain quantities to be 
1, and one of those quantities to be .8036, what is the 
other? (See Pros. 1, page 20.) 

184. Charles and Joseph together have $4.38; of 
which Charles’s share is 17 shillings and 3 pence. What 
is Joseph’s share ? 

185. Suppose .08 to be the difference between two 
quantities, and the greater quantity to be 80; what is the 

smaller? (See Pros. 11, page 20.) 

186. There is a field, 5.864 acres of which is planted 
with corn, and the rest, with potatoes. here is 2A. 
3R. 10r. more of corn than potatoes. How much is 

planted with potatoes ? 
_ 187. Suppose 7426.1 to be the difference between two 
quantities, and the smaller quantity to be .93; what is the 
‘greater? (See Pros. 111, page 21.) 
_ 188. Henry has $1.355 more money than William; 
: 6* 


, 
\ 
} 


66 ARITHMETIC. xh 


and William has 19s. 103d., New England currency, | 
How much has Henry ? 

189. What are the two quantities whose sum is 290, | 
009, and whose difference is .99? “(See Pros. IV, 
page 21. ) 

190. If a horse and chaise cost $437.25, and the) 
chaise cost $67.08 more than the horse, what is the cost 
of each? | 

191. Suppose 15675.266547 to be the produet ‘of 
some two factors, one of which is 27.381; what is the} 


other? (See Pros. v, page 21.) | 
192. If a board be 1ft. 9in. wide, how long must it 


| 

be, to contain 26.5 square feet of surface ? | 

-193. Suppose 566.916128724 to be a dividend, and 
108.273 the quotient; what is the divisor? (See Pros. 

VI, page 21.) : 

194. 4397.4 pounds of beef was equally dividéd among | 

a number of. soldiers, and each soldier received 3. 49 

pounds. How many soldiers were there? | 

195. Suppose .025 to be a divisor, and .0 x i the quo- 

tient; what is the dividend? (See Pros. vir, page 22. ] 


196. Such a quantity of bread was divided equally | 
among 13 sailors, as allowed each sailor 1.236 pounds. 
How many pounds were si 

197. If the product of three factors be 70.4597, the 
first of those factors being 3.91, and the second 3.5, what 
is the third? (See Pros. VIII, page 22.) 1 

198. What must be the depth of a pit, that is 8 ft. 5 in. 
long, and 4ft. 3in. wide, in order that it shall contain 
231 cubic feet? (Consider 231.as a product.) a 

199. Suppose the bottom of a wagon to be 9 feet long, 
and 4ft. 3in. wide; how many feet high must wood be’ 
piled in this wagon, in order that the load shall contain 
1 cord? (View the cubic feet in a cord as a product. yy 

200. Suppose wood to be piled on a base; 15 ft. 6m) 
long, and 7 ft. 9in. wide, what must be the height of the, 
pile, to contain 16 cords? 

201. Ifa stick of timber be 1 ft. 9in. wide, and 1. A ft } 
deep, what must be its length, in order that the stick 
shall contain 1 ton? | 


| ’ 2 ee . 
{ = - 


XII. 
INFINITE DECIMALS. 


—s 
, Learners, who are preparing for commercial ee and who do not intend 
prosecute an extensive course of mathematical studies. » may omit this article, 
: proceed immediately to Art. XIII. 


| INFINITE DECIMALS are those which are understood 
.0 be indefinitely continued; either by one and the same 
igure perpetually repeated, or, by some number of 
igures perpetually recurring in the same order. For 
sxample, 444444, &c. .26262626, &c. .057057057, 
Xe. 134913491349, &c# Decimals of this kind result 
rom division, when the divisor ard dividend are prime 
0 each other, and the divisor contains prime numbers 
other than those contained in 10; that is, other than 2 
id 

An infinite decimal which is continued by the repetition 


‘epeated figure is called the repetend. 

An infinite decimal which is continued by the repetition 
if more than one figure, is called a circulating decimal; 
ind the repeated period of figures is called the circ: wlate, 
wr compound repetend. ‘ee 
_ When, other decimal figures aided the repetend or 
‘irculate, the decimal is called a mixed infinite decimal. 
for example, .8476666, &c. .38171717, &c. 

_ A single repetend is ‘distinguished by a point over 1, 
lin, .8, which significa 33333, &c. A compound re- 
vetend is distinguished by a point over its first, and last 
igure, thus, .849, which signifies .849849849, &c. 
| Semilar ‘Fepetends—whether single or compound— 
re those which begin at the same place, either before or 

ter the decimal point. For example, “a and .72 are 
imilar; also, .264 and .9038 are similar; also, 3.54 and 


.86 are similar. 
| Dissimilar repetends are those which begin at different 


vlaces. For example 6127 and .405 are dissimilar. 


| 


| 
| 
1 


XU. INFINITE DECIMALS. 67 


if a single figure, is called a repeating decimal; and the | 


—* gh 


4, oF 


. 


68 | ARITHMETIC. XII 


Conterminous repetends are those which end at the 


same place. For example, .749 and .506. 
Similar and conterminous repetends are those whiel 


begin and end at the same places. For example, . 1308 
and .4012. | 


Any quotient continued by annexing decimal ciphen 
to the dividend, is known to be infinite, whenever a re. 
mainder occurs, that has occurred before; and the repe, 
tend is known to consist of those quotient figures whiel 
succeed the first appearance, and precede the secofl 
appearance of the recurring remainder. It may also he 
observed, that every quotient which does not terminate’ 
must, at some place, repeat or circulate. This truth 
evident from the consideration, that the several remain: 
ders, which precede their respective quotient figures) 
must all be within the series of numbers, 1, 2, 3, 4, anc 
so on, up to the number of the divisor. Therefore, it i 
impossible that the number of partial divisions in any 
operation shall equal the number indicated by the divisor, 
without the recurrence of some one of the remainders. | 


4 


REDUCTION OF INFINITE DECIMALS. 


CASE I. To reduce a repetend to a vulgar fraction. | 
The observations which lead to the rule are as fol 
lows. If 1, with ciphers continually annexed, be di 
vided by 9, the quotient will be Is continually; that is} 
if 4 be reduced to a decimal, it will Sy ge the repetent 
i: and since .1 is the decimal equal to 1, Q—=2,; 3 
3, A=4 , and so on, up to ios or unity. Therefotl 
every single repetend is equal to a vulgar fraction, whose 
numerator is the Tepeating figure, and whose denominato: 
is 9. Again, 1 if sy be reduced to a sete: it become: 
.01; and since .O1 is the decimal equal to J;, .02=4) 
0895, and so on, up to .99==gg or unity. Again) 
if $9, be reduced to a decimal, it pecont .001, ane 


since .001 is the decimal equal to 55, .002—=,25. 


INFINITE DECIMALS. 69 


003—-,35, and so on. This correspondence exists 
miversally; and, therefore, any circulate—not containing 
m integer— is equal to a vulgar fraction, whose numera- 
or is the circulating figures, and whose denominator is 
lenoted by as many 9s as there are places in the circulate. 
RULE. JMake the repetend the numerator, and for the 
lenominator take as many 9s as there are figures in the 
‘epetend. 
| When there are integral figures in the repetend, a 
umber of ciphers equal to the number of integral figures 
ust be annexed to the numerator. 
' 1. Reduce .6 toa vulgar fraction. 


, 2. Reduce .037 to a vulgar fraction; giving the frac- 
(On in its lowest terms. 


8. Reduce .123 to a vulgar fraction. 
4. Reduce .142857 to a vulgar fraction. 
| 5. Reduce .769230 to a vulgar fraction. 
6. Reduce 2.37 to a vulgar fraction. - 


Sey aon 


‘CASE II. To reduce a mixed infinite decimal to a 
ulgar fraction. 
Observe, that a mixed infinite decimal consists of two 
arts—the finite part, and the repeating part. The finite 
art may be reduced as shown in Art, x1, Case 1; and 
ie repeating part, as shown in the first case of this article; 
bserving, however, to reckon the value of the fraction 
btained from the repeating part ten times less for every 
lace occupied by the finite figures. For example, the 
ecimal .26 is divisible into the finite decimal .2, and the 
spetend..06. Now .2==+4,, and .6 would be=$, if 
ae circulation began immediately after the place of units; 
ut since it begins after the place of tenths, it is § of y 
Ee. “Then, .26 is equal to 2,-+¢5=3$ tos =350- 
RULE. To as many 9s as there are figures in the 
apetend, annex as many ciphers as there are finite places, 
or a denominator. Then, multiply the same number 
Qs by the finite part of the decimal, and add the repe- 
nd to the product, for the numerator. 


70 ARITHMETIC. XI 


7. What is the least vulgar fraction equal to .13? | 
8. Reduce 148 to a vulgar fraction. 
9. Reduce .532 to a ilies fraction. | 
10. Reduce .81247 to a vulgar fraction. ei 
11. Reduce .092 to a vulgar fraction. 

12. Reduce .00849713 to a vulgar fraction. 


CASE I. ‘To make any number of dissimilar rem 
tends, similar and conterminous. | 
Observe, that a single repetend may be representey 
either as a compound repetend or as a mixed decimal 
thus, .6=.666—=.66666. Also, a compound repe. 
tend may be represented as a mixed decimal; thus 
248 == .24824 = .24824824. Also, a finite decimal ma), 
be represented as a mixed infinite decimal, by annex: 
ing ciphers as repetends; thus, .39=. 300—3000=! 
390000. Hence, two or more decimals, whether repe. 
tends, circulates, or mixed decimals, may be expresse( 
with circulating figures beginning and ending together. | 
RULE. Find the least common multiple of the severa’ 
numbers of decimal places in the several repetends 
extend the repetend which begins lowest to as mam 
places as the multiple has units, and make ali the othe 
repetends to conform thereto. 
13. Make 6. 317, 3.45, 52.3, 191. 03, .057, 5.3 anc 
1.359 similar and conterminous. . 


5 tn | The first renetcrl i 
6.317= 6.31781731 3 places; the second, 2! 


3.45 = 3.45555555 | the fourth, 2; the fifth, 3) 
52.3 == 52.30000000 | the sixth, 1; the seveniil 
191.03 =191.03030303 |1. The least commor 

.057—=  .05705705 | multiple of 3, 2, 2, 3, 1, 


5.3 —= 5.33333333 tM rete therefore the simi~ 


: : lar and conterminous repe-. 
1.359 1.35999999 | tends have 6 places. 


14. Make 9. 814, 1,5, 87. 26, .083 and 124.09 sirnilae 
and conterminous., 


XII. INFINITE DECIMALS. 71 


15. Make .321, .8262, .05, .0902 and .6 similar and 
sonterminous. 

16. Make .531, .7348, .07 .0503 and .749 similar 
ind conterminous. 


| CASEIV. To find whether a given vulgar fraction is 
‘qual to a finite, or infinite decimal; and, of how many 
igures the repetend will consist. 

If we divide unity with decimal ciphers annexed 
1.0000, &c.! by any prime number, except the factors 
f 10, [2 and 5], the figures in the quotient will begin to 
epeat as soon as the remainder is 1. And since 9999, 
tc. is less than 10000, &c. by 1, therefore, 9999, &c. 
livided by any number whatever will leave 0 for a re- 
aainder, when the repeating figures are at their period. 
Now, whatever number of repeating figures we have, 
then the dividend is 1, there will be the same number, 
yhen the dividend is any other number whatever: for the 
roduct of any circulating number, by any other. given 
umber, will consist of the same number of repeating 
guresas before. Take, for instance, the infinite decimal 
386738673867, &c. whose repeating part is 3867. 
Yow every repetend [3867] being equally multiplied, 
hust produce the same product: for though these pro- 
ucts will consist of more places, yet the overplus in 
ach, being alike, will be carried to the next, by which 
reans each product will be equally increased, and con- 
2quently every four places will continue alike. From 
ese observations it appears, that the dividend may be 
tered at pleasure, and the number of places in the 
xpetend will still be the same: thus, ;,—.09, and ;3- 
ray x< 3 — D7, A 

'RULE. Reduce the vulgar fraction to its lowest terms, 
nd divide the denominator by 10, 5, or 2, as often as 
ha If the whole denominator vanish in dividing, 
e decimal will be finite, and will consist of as many 
ures as there are divisions performed. 

If the denominator do not vanish, then by the last 
uotient divide 9999, &c. till nothing remains: the num- 


“8 7 


79 ARITHMETIC. at XT 


ber of 9s used, will show the number of places in the repe 
tend; which will begin after so many places of figures a 
there were 10s, 5s, or 2s used in dwwiding. | 

17. Is the decimal equal to +, finite, or infinite— am 
if infinite, how many places has the repetend ? . 


2\i 12 Since the denominator doe 
2| 56 not vanish in dividing by 2, th 
9} 28 decimal is infinite: and, as $i 
wor 9s are used, the repetend wi 
(2) 14 consist of six figures; beginnin 


7)999999 at the fifth place, because fou. 

142857 2s were used in dividing. | 

18. Examine the fraction =4;, as above directed. 7 
19. Examine the fraction 7, as above directed. 
20. Examine ‘the fraction 7,3;, as above directed. — 
21. Examine the fraction zzz, as above directed. _ 
22, Examine the fraction 34, as above directed. 


ADDITION OF INFINITE DECIMALS. ; ) 


RULE. Jake the repetends similar and conierminom 
and add them together. Divide this sum by as many § 
as there are places in the repetend; denote the remaindt 
as the repetend of the sum, filling out its places wut 
ciphers when ti has not as many places as the repetenc 
added; and carry the quotient to the next column. __ 

23. What is the sum of 3.6-+ 78.3476 +-735.3- 
975. + .27-1 187.4? $ 

36 =  3.6666666 The sum of the repr 


rene . - | tends is first found tot 
78.3476 = 78.3476476 | 9640191, This sumis the 


735.3 =='735.3333333 | divided by 999999, and. 
375. == 375.0000000 | givesa quotient of 2, whic¢ 

One 2727272 | wecarry to the columa ¢ 
187.4 ==187.4444444 | tenths, and a remaindi 


: --—| of 648193, which we di 
1380.0645193 | note as a repetend. 


24. What is the sum of 5391.357-+-75.38-+ 187.9 
-++ 4.2965 -+ 217.8496 +-42.176-+ .523 + 58.30048 ? | 


is 
XII. » INFINITE DECIMALS. 73 


| 95. What is the sum of 9.814+1.3+-87.26 $.083 
4124.09? i + 

_ 26. What is the sum of .162 +- 134.09 + 2.93 + 
97.26-+ 3.769230 -+-99.083-++1.5+.814?, 


SUBTRACTION OF INFINITE DECIMALS. 


RULE. Wake the repetends similar and conterminous, 
and subtract as usual; observing, that, tf the repetend 
lof the subtrahend be greater than that of the minuend, 
the right hand figure of the remainder must be less by 1 
than it would be, if the expression were finite. 
| 27. Subtract 13.76432 from $5.62. 
| ee Pate Here, the whole repetend 
| 85.62 == 85.62626 of the subtrahend is greater 
—13.764382—=13.76432 | than that of the minuend, and 

¢ 71.86193 | the last figure in the remain- 
—————— | der is diminished by 1 
28. Gubiract 84.7697 from 476.32. 
29. Subtract .0382 from 3.8564. 


30. Subtract 493.1502 from 1900.842974. 


MULTIPLICATION OF INFINITE DECIMALS. 


RULE. Change the factors to vulgar fractions, mul- 
“iply these fractions together, and reduce their product 
‘0 a decimal. 


31. What is the product of 36 x .25? 
B66 — 4, 

Ii Bb 28 1X33 =390 =: 0929 Ans. 
| 32. What is the scons i 97.98 X. 262 
| 33. What is the product of 8574.3 X 87.5? 
, 34. What is the product of 3.973 X 8? 

-35. What is the product of 49640.54 X -70503 : ° 
| 36. What is the product of 3. 145 x 4.297 ? 
$7. What is the product of 8. 34.6 7.09? 
38. What is the product of .3% .09X 8.2 X.9? 

is 


74 ARITHMETIC. XIIT., 


DIVISION OF INFINITE DECIMALS. 


RULE. Change both divisor and dividend into vulgar 
fractions, find their quotient in a vulgar fraction, and 
reduce it to a-decimal. | 

39. What is the quotient of .36 by .25? 

36= =, 

Bak kt R= eX 

Then, 132% —=1.422924901 1857707509881 Ans. 
40. What is the quotient of 234.6 by .7? | 
41. What is the quotient of 13.5169533 by 4.297? 


42. Divide 319.28007112 by 764.5. | 


| 
| 


XIII. 
RELATIONS OF NUMBERS. 


"Any number may be viewed as a part, or as so many 
parts of any other number ; and it is in this view, that wr 
shall, at present, notice the RELATION of one number t 
another. 

For example, 1 is } of 5, 3 is 3 of 5, 9 is 2 of 5, &e, 
Here 1 stands in the same relation to 5 that 4 does to: 
unit; 3 stands in the same relation to 5 that 2 does to) 
unit; and 9 stands in the same relation to 5 that 3 doe) 
to aunit. Thus, the number which is viewed as the par, 
or parts of another, becomes a numerator, and the othe: 
number the denominator of a vulgar fraction. This frac 
tion may be reduced, and the relation it expresses wi 
remain unaltered. For instance, 4 of 8 is the same as~ 
of 8; and 24 of 8 is the same as § or { of 8. 

In the various practice of arithmetic, most of the sola 
tions are performed by process to which the performer 1 
led, by considering the relation which exists between th 
numbers concerned. The truth of this remark will ap 
pear evident to the learner, in the course of subsequen 
exercises. | 


‘XII. RELATIONS OF NUMBERS. 15 


1. Express 16 as a fractional part of 56, and reduce 
the fraction to its lowest terms. 
2. Express 9 as the fractional part of 45, and reduce 
the fraction to its lowest terms. 

3. What part of 34 is 20? What part of 34 is 21? 

4. What part of 34 is 49 ?—Or, in other words, what 
is the improper fraction that expresses the relation in 
which 49 stands to 34? 

5. What part of 24 is 36? What part of 24 is 37? 


| 6. What part of 2 yards 1ft. 6in. is lyd. 2ft. 10in.? 
In this example, 2yd. I ft. 6in. becomes a denomina- 
tor, and Lyd. 2ft. 10im. the numerator. But both these 
quantities must be reduced to their lowest denomination, 
inches; the relation will then be simple, and may admit 
of being reduced to lower terms. 
7. What part of 1 yard is 2 feet 6 inches? 
8. What part of £3 14s. is 16s. 10d.° 
9. What part of 9s. 7d. 2qr. is 2s. 9d. Iqr.? 
10. What part of 5 gallons 2 pints is 3 quarts 3 gills? 
1]. What part of 2 acres is 1 acre 3 roods 32 rods? 


12. What part of $7 is $4.65? 
$7700 cents, and $4.65==465 cents. Then 465 
cents is 463 of 700 cents. 72225. 
When either or both the numbers, whose relation is to 
be expressed, contains a decimal fraction, the decimal 
places i in the two numbers must be made equal —if they 
are not already so—by annexing decimal ciphers. The 
decimal points may then be erased, and the numbers 
written as the terms of a vulgar fraction. For example, 
the relation of .14 to 9 is }4,—= i. 
13. What part of 2.1 is 1.72? 
14. What part of 4.87 is 2? 
15. What part of $24.08 is $15: > 
| 16. What part of .65 is .408° 
_. 17, What part of $2 is $7? (Ans. 3.) 
18. What part of $2 is $7.49? | 
19. What part of 90 cents is $1.35? 
20. What part of $4.375 is $28? 


7 ARITHMETIC XI 


21. What part’of 5.8 is 31.42? 
22. What part of .253 is .97? 


23. What part of $ is 75? | 

The expression oF a4 sa oh Is, at first, a complex 
fraction, of which ; is the numerator, and 4 the ieatcal 
nator. The expression may be simplified by reducin 
these fractions to a common denominator, and taking th 
new numerators for the terms of the relation. See a 
to reduce a complex fraction to a simple one, page 43. 

24. What part of 12 is sin 

25. What part of 3 si 

26. What part of 7 Is $ 

27. What part of 6} is ter P 

28. What part of 18 is 23 > 

29. What part of 24 feet is 102 inches? 

30. What part of 145 days is 23.1, 5 hours ? 

31. What part of 24 “gallons i is 3 quarts 23 gills? @ 

32. What part of 54 rods is 3 rods 22 ft.? 4 

33. What part of 4 1s 3? : 

34. What part of $is $? 

35. What part of 3 7. ions 

36. What part of $3 * 96: 

37. What part of 1g is 3? | 

38. What part of 3 ‘shillings i is 5s. 7d.? - i 

39. What part of £1 14s. is £5 2s. 73d.? : 

40. What part of 784 days is 125 days 178 hours? 

41. What part of 22 tons is 4 tons 64 pounds? i 


42. If 35 horses eat 12278 pounds of hay in a wee 
what will 17 horses eat, in the same time ? 4 

The most obvious view of the solution of this question | 
is this —If 35 horses eat 12278 pounds, 1 horse will eat | 
zs Of 12278 pounds, which is 35033 pounds; and 17 }| 
horses will eat 17 times 35033 pounds, which is 623 
pounds. A more concise view, however, may be taken, | 
as follows. 17 horses are 42 of 35 horses, and the 
will eat 32 of the 12278 pounds of hay. Therefore, we 
shall obtain slic answer by multiplying 12278 pounds by 
the fraction 3. 12278X 346230 Ans. | 


: 


eT. RELATIONS OF NUMBERS. ror 


43. If a car run 552 miles upon a rail-road, in 24 
hours, how far will it run in 13 hours ? 
44. If a car run 3 miles [960 rods] in 38 minutes [480 
seconds], in what time will it run 300 rods? 
45. If a hogshead of wine [63 gallons] cost $98.50, 


_ what will 45 gallons cost, at the same rate? 


46. If the annual expense of supporting a fort manned 


_ with 600 soldiers be $ 182571, what is the exneng of 
a fort manned with 424 soldiers ? 


47. If I can buy 325 barrels of flour for $1425, pave 


_ many barrels can I buy for $521? 


48. Ifa ferry boat cross the river 18 times in 5 ea 
in how many hours will it cross 4 times ? 


49. If 9barrels of flour cost $ 32, what will 28 bl. cost? 

In this example, the relation in which 28 barrels stand 
to 9 barrels is expressed by an improper fraction; 28 
barrels being *° of 9 barrels. Therefore the answer is 
obtained by multiplying $32 by 32; that is, by multiply- 
ing $32 by 28, and dividing the product by 9. 

50. If it take 300 yards of cloth to make the uniform 
clothes for 52 soldiers, how many yards are required to 
clothe 784 soldiers ? 

51. If 12 horses eat 20 bushels of oats in a week, 
how many bushels will 45 horses eat in the same time ? 

52. Ifa post 5 feet high cast a shadow 3 feet, on level 


-ground, what is the height of a steeple, which, at the same 


time, casts a shadow 176 feet? 

53. If $40 will pay for 144 yards of cloth, how many 
yards can be bought for $75 ? 

54. If 95 bushels of corn cost $68.25, what will 320 
bushels cost, at the same rate ? 

55. Suppose a ship’s expenses in Liverpool to be 
£131 13s. 10d. for 22 days; what-would be her ex- 
penses in the same port for 35 days? 

56. If 144 bushels of corn will grow upon 38 acres 1 
rood 15 rods of land, how much land is necessary to 
produce 500 bushels ? 

57. Bought 269 yards of cloth, at the rate of $ 100 for 


_ 30 yards. What did it amount to ? 


7% 


ae 
» 


18 ARITHMETIC. XII, } 


58. Bought 24yd. Sqr. Ina. of cloth, at the rate of | 
$12.30 for 4yd. Igr. 2na. What did it amount to? 9) 
Since it is necessary, in this example, to consider 24 | 
yd. 3qr. Ina. as a fractional part of 4 yd. I qr. 2na., they) 
first step in the operation is, to reduce both quantities of | 
cloth to nails. , 
59. If 13gal. 2 at. 1pt. of wine cost poe 16, what 
bi sk gal. 3 qt. 1 pt. cost, at the same rate? 4q 
If 26 barrels of flour cost £ 28 14s. 6d. how many 
oe will £35 10s. 4d. pay for? q 
61. If 6gal. Qqt. 1 pt. of wine will fill 31 bottles, how) 
many bottles are required for 11 gal. 3 qt.? 4 
62. If 144 gross of buttons cost £22 19s., how many 
gross can be bought for £12 5s. 54d.? 
63. If 2hhd. 19 gal. 2qt. of wine cost £93 1s. a 
what will 25hhd. 36 gal. cost? 
64. If 15 yards of cloth cost $39.45, how many rcs 
can be bought for $21? (See remark ciel example 12.) 
65. At the rate of # 94 for 78 days’ work, in how many 
days can a labourer earn $'72.375 ? 
66. At the rate of ¢ 240 for 9.5 acres of land, what is 
the value of 7.25 acres? 
67. Atthe rate of $182.50 for 8 acres of land, wall is 
the value of 12.7 acres? | 
68. At the rate of 75 cents for 92 of a bushel of corn, 
what is the, value of .648 of a bushel ? | 
69. In how many minutes will a locomotive car rum 
49.9 miles ; il pedine: it to run at the rate of 2.5 miles in 
5.75 minutes ? 7 
70. If 43.64 pounds of satis be worth $9.075, whale 
is the value of 108.9 pounds? oe} 
71. If 14 dollars will pay for the carriage of a ton 75.6 e | 
miles, what distance can a. ton be carried for 16 dollars | 
75 cents, at the same rate ? 


72. If 4 of a yard of cloth cost $7, Be is the cost | 
of ; of a yard? (Recur to example 23.) 
73. If a rail- road car run 260 miles in 12 hours, wha 
distance will it run in 102 hours? (See example 24.) 


} fe 
CHI. RELATIONS OF NUMBERS. 79 


| 74. If a man earn ol. 15 in 2 of a day, how much can 
e earn in $ of a day? (See example 25.) 

m5. Suppose 3 of an acre of land to be worth 54 dol- 
us; what is > ot an acre worth ? 

“To solve this question, by the eee to which the 
cholar has been led, he will consider ¢ as a denominator 
ad 4 as the numerator of a complex fraction, expressing 
‘hat part of 54 dollars 3 of an acre is worth; and, after 
bdacin this complex fraction to a simple one, will mul: 
ply the simple fraction into 54 dollars, for the answer. 
low the effect of the process is the same as that of mul- 
plying the 54 by 5, and dividing the product by 4; and 
us last method is to be preferred, because it is shorter. 
“hus, 54x96, and rina Deen 

76. If of aship cost $15000, what does 4 L of tier cost? 

ay, Iii i of a lot of new piletnd be worth 300 ialiinss what 

45 of the lot worth? 

78. If a horse trot 1840 rods in 49 of an hour, how 
rany rods does he trot in 3% of an hour? 

79. If 964 yards of cloth cost $642, what will 28 
ards cost, at the same rate ° 

80. If 154 yards of cloth cost $75, what will 142% 
) ost, at the same rate ? 

81. If 97 barrels of flour be consumed by a company 
118 days, how long will 253 barrels last ? 

82. Ifa mill grind 18;% bushels of corn in | hour and 
2 minutes, in what time val it grind 253 bushels ? 

83. Ifa ship sail 924 miles in 83 hours, j in how many 
ours does it sail 65,°, miles ° 


- 


84. If a barrel of flour will support 12 men for 25 
ays, how long will it support 8 men? 

Since the flour will support 12 men 24 days, it would 
apport 1 man 12 times 25 days, or 300 days; and since 
would support 1 man 300 days, it will support 8 men 
of 300 days, or 874 days. Thus, to obtain the answer, 
e multiply 25 days by 12, and divide the product by 8. 

. little attention to the conditions of. this question, and 
le process of the operation, will enable the learner to 


erceive, at once, that the answer is 1 of 25 days. 


86 ARITHMETIC. XID) 


85. If a quantity of beef will support 436 men 73 days 
how long will it support 240 men? 

86. Ifa barrel of beer will last 10 men 16 days, hoy 
long will it last 23 men? 

The beer would last 1 aaa 10 times 16 daysye or 6 
ee and it will last 23 men 35 of 60 days, or 244 day 
=—=2,/, days. The question” is, however, more conve 
niently viewed thus,—Since the beer will last 10 me 
16 days, it will last 23 men? a3 Of 16 days; and, hence 
16 is to be multiplied by 38 

87. Suppose a certain quantity of hay will feed 8 
sheep 71 days; how long will it feed 230 sheep ¢ 

88. If 256 men can make a certain piece of road i 
240 days, in what time will 190 men make it? 

89. If 9 yards of silk, that is 3 cieeist wide, will lin 
a cloak, how many yards, that is 5 quarters wide, wi 
line the same cloak ? | 

90. If 110 yards of paper, that is 52 inches wide, wi 


cover the walls of a room, how many yards; that is 2 


ae wide, will cover the same walls ? 
Suppose a man can perform a piece of work | 


5 
' 


45 ay s, by working 7 hours a day, in what time will h_ 


perform it, if they work 10 hours a day ? 

92. Suppose a company of men can perform a piec 
of work in 155 days, by working 12 hours a day, in whe 
time will they perform it, by working 5 hours a day? 

93. How many days will it take 119 horses to eat th 
hay that 44 horses would eat in 60 days ? 

94. The hind wheels of a coach, which are 180 inche 


in circumference, will turn round 4825 times in runnin 


acertain distance, how many times will the forward wheel 
turn round, they being 145 inches in circumference ? 
95. If a ship, by sailing 9 miles an hour, will effect — 


passage to Europe in 55 days, in how many days woul 


she effect the passage by ples 13 miles an hour? 
96. Ifa vessel, by sailing 10} miles an hour, will mak 


a passage from Bangor to New Orleans in 11 days, 1. 


how many days would she make the passage by sain 
12! miles an hour ? 
97. Suppose A rides 62 miles an hour, and perform 


XIII. RELATIONS OF NUMBERS. St 


a certain journey in 14°; days; in what time will B; who 
rides only 4,’; miles an hoor; perform the same journey? 


98. If 6 persons expend $300 in 8 months, how much 
will serve 15 persons for 20 months? 

Since 6 persons expend $300 300 750 
in 8 months, 15 persons would, 15 20 
in the same time, expend 'P of a 
$300, which is 4750. Then, 6)4500 — 8)15000 
since 15 persons would expend 750 1875 
$750 in 8 months, they would, in 20 months, expend 
* of $750, which is $1875. The adjoined operation 
corresponds to this solution. | 

99. If the wages of 6 men for 14 days be $84, what 
will be the wages of 9 men for 11 days? 
- 100. If 3 pounds of yarn make 9 yards: of cloth, 5 
quarters wide, how many pounds would be required to 
make a piece of cloth 45 yd. long and 4qr. wide? 
- 101. If aclass of 25 girls perform 1750 examples in 
arithmetic, in 15 hours, how many examples of equal 
length may a class of 30 girls perform, in 18 hours ? 
| 102. If the use of $100 for 90 days, be worth $1.50, 
what is the use of $73 for 85 days worth? 
103. If the use of $ 100 for 30 days be worth 75 cents, 
what is the use of $1240 for 57 days worth ? 

104. If a man euvel 217 miles in 7 days, travelling 6 
hours a day, how many miles will he travel in 9 days, if 
he travel 11 hours a day? 
| When he travels 6 hours a day, he divéiees 217 miles 
i 7 days, and were he to proceed thus for 9 days, he 
would advance 7 of 217 miles, or 279 miles. Since, by 
travelling 6 hours a day he would, in 9 days, advance 
279 ml., ~ by travelling 11 hours a day , he would advance 
Ksof 279 ml., which is 5112 ml., or 5115 ml. 

105. If a man perform a aounien of 1250 miles in 15 
lays, by travelling 14 hours a day, how many days will 
it take him, to per phen a journey of 1000 miles, by travel- 
ling 13 hours a day? 

106. If 10 cows eat 7} tons of hay ii in 14 weeks, how 
imany cows will eat 224 tons in 28 weeks ? 


| 


S2 ARI'FHMETIC. XID 


eB) 


107. If 6 men will mow 35 acres of grass in 7 days 
by working 10 hours a day, how many men will be re 
quired to mow 48 acres in 5 days, when they work 1 
hours a day? | 

108. If 14 men can cut 87 cords of wood in 3 days | 
when the days are 14 hours long, how many men will ct 
175 cords, when the days are 11 hours long ? 

109. If 16 men can build 18 rods of wall in 12 days. 
how many men must be employed to build 72 rods of th’ 
same kind of wall in 8 days? | 

110. If 25 persons consume 600 bushels of corn i 
2 years, how much will 139 persons consume in 7 years” 

Since 25 persons consume 600 bushels in 2 years 
139 persons would, in the same time, consume 4% 
600 bushels, which is 3336 bushels. Then, since 13) 
persons would consume 3336 bushels in 2 years, the’ 
will, im 7 years, consume 4% of 3336 bushels, which 1) 
11676 bushels. . 

111. If 154 bushels of oats will serve 14 horses fet 

14 days, how long will 406 bushels serve 7 horses? | 

112. If 25 men can earn $6250 in 2 years, how lon) 
will it take 5 men to earn $11250? | 

113. If 9 men can mow 36 acres of Soa ls in 4 days | 
how many acres will 19 men mow in 11 days ? i. 

114. Ifa family of 9 persons spend $450 in 5 months” 
how much would be sufficient to maintain the family 
months, after 5 more persons were added ? | 
_ 115. Ifa stream of water running into a pond of 19) 
acres, will raise the pond 10 inches in 12 hours, hov 
much would a pond of 50 acres be raised by the sam 
stream, in 10 hours? 

116. If the wages of 4 men, for 3 days, be $11.04 
how many men may be hired 16 days for $103.04? 
117. If 3 men receive £8 18s. for working 194 day: 
what must 20 men receive for working 1004 days: ; | 

118. If 1112 bottles are sufficient to receive 5: casks 
of Mads. how many bottles are sufficient to receive 1¢ 
casks of wine? 

119. If 725 bottles hold 4 barrels of wine, how many 
notiles are required to hold 3 tierces of wine ? a 


| 


} 


XM. RELATIONS OF NUMBERS. $3 


120. If 240 men, in 5 days, of 11 hours each, can dig a 
trench 230 yards long, 3yards wide, and 2 yards deep, in 
how many days, of 9 hours each, will 24 men dig a trench 

420 yards long, 5 yards wide, and 3 yards deep ? 

Since 248 men, in 5 days, of 11 hours each, can dig 
a trench 230 yards long, 3 yards wide, and 2 yards deep, 
24 men, working in days of the same length, would dig a 
trench of the same dimensions in 248 of 5 days, which is 

§13$—514 days; and, working in days of 9, instead of 
11 hours each, the trench would occupy them '¥} of 514 
days, which is 633, days. Again, since the trench to be 
dug by 24 men is 420, instead of 2350 yards long, this 
length, (the width and depth remaining unchanged) would 
occupy them $39—33 of 633 days, which is 115%. 
days. Again, since the trench to be dug by 24 men is 
5, instead of 3 yards wide, this width (the depth remain- 
ing unchanged) would occupy them 3 of 115%, days, 
which is 1922338 days. Lastly, since the trench to be 
dug by 24 men is 3, instead of 2 yards deep, it will 
occupy them 3 of 192434 days, which is 288355— 2883, 
days, the answer. 

121. If12 men can build a brick wall 25 feet long, 7 feet 
high, and 4 feet thick, in 18 days, in how many days will 
20 men build a brick wall 150 feet long, 8 feet high, and 
5 feet thick ? 

122. If 15 men can dig a trench 75 feet long, 8it. 
wide, and 6ft. deep, in 12 days, how many men mus? 
be employed to dig a trench 300ft. long, 12 ft. wide, and 
9ft. deep, in 10 days? 

123. If the carriage of 44 barrels of flour, 108 miles 
be worth $215, what is the carriage of 36 barrels, 162 
miles worth ? . : 

124. If 175 bushels of corn, when corn is worth 60 
cents a bushel, be given for the carriage of 100 barrels 
of flour, 58 miles, how many bushels of corn, when corn 
is worth 75 cents a bushel, must be given for the carriage 
of 90 barrels of flour, 200 miles ? 

125. If 12 ounces of wool make 25 yards of cloth, 
that is 6 quarters wide, how many pounds of wool would 
make 150 yards of cloth, 4 quarters wide ? 


84 ARITHMETIC. XII 


MISCELLANEOUS EXAMPIES. 


126. A owned 3 of a ship, which he sold for $ 365¢ 
and B owns -%; of her, which he wishes to sell at the sam 
rate. What must be B’s price? 

Since the price of 3% of the ship is $3650, the pric 
of the whole ship must be % of § 3650, which is $ 36906 
and +5 of $36900 is $11070, which must be B’s price 

127. If 3650 be 3 of some number, what is 33; of th 
same number ? y 

128. A merchant has bought 4 of a company’s stock 
for $92000. What would be the price of <4; of the stock 
at the same rate ? | 

129. A merchant owning 3/5 of a ship, sold 2; of whe 
he owned for $1841. What is the value of the whol 
ship, according to this sale ? | 

130. 1841 is ~% of 4% of what number ? 

131. After a certain tract of land had been equall 
divided among 16 owners, one of them sold 2 of his shar 
at $5, an acre, and received $444. How much land wa’ 
there in the whole tract ? 

132. If 4% of a yard of cloth be worth 2 of a dollar 
what is the value of -2; of a yard? | a 

Since 7% of a yard is worth = of a dollar, a yardi 
worth 1 of 2 of a dollar, which is 22 of a dollar; ant 
¢z of a yard is worth {3 of £2 of a dollar, which is 7 
of a dollar, or 61328 — ¢ 143 — 1.5594. | 
_ 183. If § of a yard of lace be worth 12 of a dollar 
what is +2, of a yard worth? | 

134. If 4 of a barrel of flour cost 4 dollars, what 1: 
the cost of 63 barrels, at the same rate? —- | 
135. If 13% bushels of corn cost 7 dollars, what i: 
. the price of 93 bushels, at the same rate ? ¢ 
_ 136. If 422 pounds of indigo be worth $87.625, wha’ 
is the value of 1923 pounds ? ¥ , 

137. A garrison of 900 men have provision for 4 
months. How many men must leave the garrison, that 
the provision may last the remainder 9 months ? : 

138. Ifa loaf of bread weighing 32 ounces be sold for 


eight cents, when flour is worth $6.50 per. barrel, what 
Y ; 


XIII. MISCELLANEOUS EXAMPLES. 85 


ought the eight-cent loaf to weigh, when flour is worth 
only $5 a barrel ? 
' 1389. Acompany of 75 soldiers are to be clothed; each 
_ suit is to contain 33 yards of cloth, 6 quarters wide, and 
to be lined with flannel 3 of a yard wide. How many 
_ yards of flannel will be required ? 
_ 140. If a garrison of 1500 men consume 750 barrels 
of flour in 9 months, how many barrels. will 2150 men 
consume in 15 months? 
_ 141. How many tiles 8 inches square, will cover a 
hearth 16 feet long, and 12 feet wide? 
» 142. If the expense of carrying 17cwt. 3qr. 14lb. 

85 miles be $23.84, what will be the expense of carrying 
S3cwt. 2qr. 150 miles, at the same rate? 

143. Two men bought a barrel of flour; one paid 3} 
dollars, and the other paid 37 dollars. What part of the 
flour should each of them have ? 

_ 144. If the corn contained in 8 bags, holding 2 bushels 
‘3 pecks each, be worth $14.25, what is the value of the 
corn contained in 7 bags, each holding 2bu. 3pk. 7 qt? 

145. A ship of war sailed with 650 men, and provision 
for a cruise of 15 months. At the end of 3 monihs she 
captured an enemy’s vessel, and put 75 men on board of 
her. Five months after, she captured and sunk another 
vessel, and took on board ihe crew, consistmg of 350 
men. How long did the provision last, from the com- 
mencement of the cruise ? 

146. A built 156 rods of wall in a certain time, and B 
in the same time built 13 rods to every 12 that A built. 
‘They were paid $1.25 per rod. How much did B re- 
-Ceive more than A? 

_ 147. A father bequeathed $6000 as follows; viz. 2 to 
his wife, 4 to his.son, + to his daughter, and the remain- 
der to his servant. How much did each receive? 

148. If 42 of a pound of sugar be worth 2 of a shilling, 
what is the value of ¢ of a cwt.? 

149. If 754% gallons of waaay one hour, run into a 
cistern, which will hold 64 hogsheads, and by a pipe 244 
gallons an hour run out, in how many hours, minutes and 
seconds will the cistern be flled ? 

8 


\’ 
bel 


® 


? 
“ 


86 ARITHMETIC. XIV.) 


XIV. 


PERCENTAGE. | 


Under this head may be classed, those computations! 
which investigate the value of a given number of hun-) 
dredths of any quantity. The number of hundredths to 
be taken or considered in any number, is called the per 
cent. The term, per cent., is an abbreviation of pet) 
eentum, which signifies by the hundred. 

Any per cent. is conveniently expressed bya decimal 
Thus, 1 per cent. of any number is .01 of that aeiitiks | 
8 per cent. is .08; 25 per cent. is .25; &c. 


1. A merchant, who has 426 dollars deposited in tH 
bank, wishes to draw out 5 per cent. of his deposite.: 
How many dollars must he draw ? 


Since 5 per cent. of any quantity is zjy of 1426 
that quantity, the ee to be solved in this 05 
example is— What is +2, of 1426dollars? Or, $71.30! 


decimally— What is .05 of 1426 dollars? The 
answer is conveniently found by multiplying 1426 by .05._ 
The whole number in the product expresses dollars, ane 
the decimal expresses cents. 

2. A trader, who went to the city with 321 dollars, to 
purchase goods, laid out 9 per cent. of his money for. 
coffee. _ How many dollars did he pay for coffee ? | 

3. What is 1 per cent. of 100 dollars ?* | 
. What is 1 per cent. of 834 dollars ? 


OO NED ov 


What is 3 per cent. 
What is 3 per cent. 
What is 6 per cent. 
What is 6 per cent. 
What is 7 per cent. 


of 100 dollars ?. 
of 42 dollars ? 
of 100 dollars ? 
of 99 dollars ¢ 

of 100 dollars ? 


10. What is 7 per cent. of 1000 dollars ? 
11. What is 8 per cent. of 26 dollars ? 

12. What is 9 per cent. of 354 dollars ? 
13. What is 10 per cent. of 2244 dollars ? 


_ XIV. PERCENTAGE. 87 


14. What is 16 per cent. of 13 dollars ? 

15. What is 37 per cent. of 211 dollars? 

16. What is 99 per cent. of 100 dollars ¢ 

17. What is 100 per cent. of 48 dollars ° 

18. A trader laid out 1214 dollars as follows. He 
paid 24 per cent. of the money for broadcloths; 38 per 
cent. for linens; 8 per cent. for calicoes; and the remain- 
der for cottons. How many dollars did he pay for each 
_kind of goods? 


When the rate per cent. is a vulgar fraction, or a mix- 
ed number, the fraction may be changed to a decimal. 
Observe, that, 1 per cent. when expressed decimally, is 
.01; therefore a fraction of 1 per cent. when reduced to 
a decimal, becomes so many tenths, hundredths, &c. of a 
pundredth. For example, as 4 of 1 unit is .25 of a unit, 
so of 1-hundredth is .25 of a “hundredth, and is denoted 


thus, .0025. 
19. What is 34 per cent. of 243 dollars? 
3 per cent. —=.03 243 
5 per cent. —.005 -035 
.035 1215 
729 


8.505 Ans. $8.504 


20. What is 44 per cent. of 2746 dollars ? 

21. What is 74 per cent. of 41 dollars ? 

22. What is 123 per cent. of 358 dollars ° 

23. What is 2 per cent. of 100 dollars ° 

24, What is 2 per cent. of 61 dollars? 

25. What is 1 per cent. of 9487 dollars ? 

26. If 84 per cent. be taken noe 36 dollars, how 
many dollars will there be remaining ? 

27. A merchant who had 400 barrels of flour, shipped 
421 per cent. of it, and sold the remainder. How many 
barrels did he sell ? 

28. A trader bought 800 pounds of coffee; and, in 
getting it to his store, 24 per cent. of it was wasted. 
How many pounds did he lose?» What did the remainder 
amount to, at 13 cents a pound? 


88 ARITHMETIC. Bs XIV) 


29. Two men had 120 dollars each. One of them 
paid out 14 per cent. of his money, and the other 17%) 
per cent. How many dollars did one pay more than the | 
other ? | 


30. Find 74 per cent. of $344. | 
When there is a fraction in the 344+ 100—3. a4 } 


rate per cent. which cannot be 74 
exactly expressed by a decimal— z 

: 2408 | 
as in this example— we first find 1142) 
1 per cent. of the given sum, by 5 


dividing it by 100; that is, by cut- $25.22) 
ting off two decimal figures, and then multiply this quo- 
tient by the mixed number expressing the ie per cent. 
31. What is 44 per cent. of 624 dollars ? | 
32. What is 62 per cent. of 38 dollars? d 
33. What is 34 per cent. of 2310 dollars? i 
34. What is 94 per cent. of 17 dollars? 
35. What is 83 per cent. of 152 dollars? ) 
36. Find the difference between 52 per cent. of 41 | 
dollars, and 44 per cent. of 39 dollars. 


37. What is 7 per cent. of $24.32? 

Here we have cents { decimals] in the num- 24.32 
ber on which the percentage is to be taken. 07 | 
We however multiply as usual in decimal ae 
multiplication; ; and the first two decimal fig- 1.7024 
ures in the product express cents, the third figure ex- 
presses mills, and the fourth expresses tenths of a mill. 

38. What is 14 per cent. of $641.94 ? 

39. What is 44 per cent. of $37.26? 

40. What is 1i2 percent. of $150.75 ? 

41. What is 124 per cent. of $25.32? 

42. If a horse and gig cost 400 dollars, and the gig 
cost 32 per cent. of the sum, what did the horse cost? — | 

43. Find the difference between 131 per cent. of 
. $18.09, and 7 per cent. of $41. 

44. Find the difference between 9 per cent. of $ 16, | 

and 84 per cent. of $17.30. 


oo) 


— —— SS 


exIV. PERCENTAGE. 89 


45. A young man, who had 94 dollars deposited in the 
Savings Bank, drew out 25 dollars. What per cent. of 
his deposite did he draw out ? 

We perceive, that the sum 94)95.0(.96 54 —96 24 
he drew out, was 23 of the sum inte Salih ik} 
he had deposited: and, since PaaS 
the rate per cent. of any sum ree 


is a certain number of hun- Sree 


paid ? 


dredths of that sum, the ques- 56 


tion to be solved is— How many hundredths is 25- 


ninety-fourths >— To solve this question, we change 25 
Ss, q ) £ 4 


to a decimal; restricting the decimal to hundredths; that 


is, carrying the quotient no further than two places. 
Any remainder which might allow the quotient to be 
carried further, may, in cases like this, be expressed in a 
vulgar fraction. Ans. 2644 per cent. 
46. A man, who was owing a debt of 240 dollars, has 
paid 32 dollars of it. What per cent. of the debt has he 


47. A merchant gave his note for 235 dollars, and 
soon after paid 110 dollars of the sum. What per cent. 
did he pay; and what per cent. still remamed due ? 

48. If the cloth for a coat cost 12 dollars, and the 
making 7 dollars, what per cent. of the whole expense is 
the making ° 

49. What per cent. of 100 dollars is 6 dollars ? 

50. What per cent. of $28.50 is $1.10? 

51. What per cent. of $94.12 is $4.42? 

52. What per cent. of $57.08 is 32 cents ° 

53. What per cent. of $10.10 is 7 cents ? 

54. What per cent. of $48.11 is 99 cents ? 

55. What per cent. of $75 is $4.18 ? 


To find the value of a rate per cent. on any sum of 


‘English money,— First, change the lower denominations 


of money in the sum, to a decimal of the highest denomi- 

nation; and then proceed to multiply by the rate, as if 

the sum were dollars and cents. The whole number in 

the product will be of the same denomination of money . 

with the whole number in the multiplicand; and the 
8 * 


‘ 
om ' 


90 ARITHMETIC. XIval 


decimal in the product must be changed to the lower 
denominations. 4 
56. An English gentleman took passage from Liver- 
pool to Boston, in the ship Dover, having £672 12s. 
4d. He paid 5 per ane of his funds for his passage. 


How much did he pay ? a 


12) 4. .6308 
20|12.333-- 20 
672.616 _ 12.6160 

; 05 12 
33.63080 7.3920 
4 


~ 3.680Ans. £33 19s. 7d. qr. + | 


57. What is 8 per cent. of £47 18s. 7d.? 
58. What is 3 percent. of £9 14s. 3qr.? 

59. What is 16 per cent. of £22 16s.? 

60. What is 25 per cent, of 19s. 8d. 2qr.? 
61. What is 6 per cent. of £2584? 

62. What is 50 per cent. of 18s. 10d. 2qr.? 
63. What is 44 per cent. of £214 15s. 10d.? 


COMMISSION. 7 

ComMISssION is the compensation made to factors and 
brokers for their services in buying or selling. It is 
reckoned at so much per cent. on the money employed 
in the transaction. 


64, What is the commission on £500 at 2} per cent.? 
65. Suppose I allow my correspondent a commission 


of 2 per cent., what is his demand on the disbursement — 


of £369? 

66. If I allow my factor a commission of 3 per cent. 
for disbursing £748 11s. 8d. on my account, what does 
his commission amount to ? 

67. How much does a broker receive on a sale of 

«stocks amounting to 52648 dollars, allowing his commis- 
sion to be 4 of 1 per cent.? 


XIV. STOCKS. 91 


68. What is the amount of commission on 395 dollars 
75 cents, at 33 per cent.? 

69. A commission merchant sold goods to the amount 
of 6910 dollars and 80 cents, upon which he charged a 
commission of 24 per cent. How much money had he 
to pay over to his employer ? 

70. Sold 94 tons, 17cwt. 3qr. of iron, at 96 dollars a 
ton, at a commission of 24 per cent. on the sale. What 
did my Bominianott amount to? How much had I to 
pay over? 


STOCKS. 


STOCK is a property, consisting in shares of some es- 
tablishment, designed to yield an income. It includes 
government securites, shares in incorporated banks, in- 
surance offices, factories, canals, rail-roads, &c. 

The nominal value, or par value of a share, is what it 
originally cost; and the real value, at any time, is the 
sum for which it will sell. When it will sell for more 
than it originally cost, it is said to be above par, and the 
excess is stated at so much per cent. advance. When 
its real value is less than the original cost, it is said to be 
below par, and is sold at a discount. 


71. Sold 10 shares in the Commonwealth Insurance 
Company, at 5 per cent. advance, the par value of ashare 
being 100 dollars. How much did I receive ? 

72. Bought 15 sharesin the Boston Bank, at 3 of 1 
per cent. advance, the par value being 50 dollars a share. 
How much did I give for them? 

73. Sold 64 shares in the State Bank, at 14 per cent. 
advance, the par value being 60 dollars a share. How 
much did I receive for them 

74. Sold 9000 dollars United States 5 per cent. stock, 
t an advance of 74 per cent. What was the amount of 
the sale ? 

_ 75. Sold 18 shares in an insurance office, at 12 per 
ent. discount, the par value being 100 dollars a share. . 
How much did they come to? 


| 
) 


76 Bought 16 shares in the Massachusetts Bank, at 1, 
per cent. advance, the par value bemg 250 dollars. i 
share. What was the amount of the purchase? 

| 
it 


92 ARITHMETIC. “ 


77. Bought 54 shares in the New York City Bank, a. 
74 per cent. advance, the par value being 100 dollars ; / 
snare. How much did they cost me? | 

78. I directed a broker to purchase 25 shares of rail 
road stock, at a discount of 13 per cent. the par valu 
being 100 per share. Allowing the broker’s commissioi 
to be ~ per cent., what will the whole cost me? 

79. What will 16 shares in the Philadelphia Bank cost 
the par value being $100 per share, the price being 3; 
per cent. above par, and the broker charging a commis 
sion of $ per cent.? 


INSURANCE. 


INSURANCE is security given, to restore the value o/ 
ships, houses, goods, &c., which may be lost by the 
perils of the sea, or by fire, &c. The security is | 
in consideration of a premium paid by the owner of the 
property insured. 

The premium is always a certain per cent. on the value 
of the property insured, and is paid at the time the msurj 
ance is effected. | 

The written instrument, which is the evidence of the 
contract of indemnity, is called a policy, 


80. What jis the amount of premium for insuring) 
19416 dollars at 25 per cent.? 
81. I effected an insurance of 3460 dollars on my 
dwelling house for one year at 2 of 1 per cent. What 
did the | premium amount to ? | 
82. If you obtain an insurance on your stock of zodid 
valued at 7325 dollars, at} of 1 per cent. what will the 
premium amount to ? it 
83. If you should take out a policy of 3168 dollars, 
on your store and goods, at a premium of 41 cents on a 
hundred dollars, what would be the amount of premium / 
84. An insurance of 18000 dollars was effected on 


| 
| 
| 
1 


xv. INTEREST. 93 

Ibe ship Sturdy, on her last voyage from Boston to Cal- 

‘utta, at a premium of 3 per cent. out and home. What 

ied the premium amount to ? 

} 85. An insurance of 3500 dollars on stock in a cotton 
actory was effected at 31 per cent. for one bab What 
was the amount of premium? 

' 86. A gentleman procured an insurance for one year 
mm his house valued in the policy at 8756 dollars, and on 
is furniture valued at 2139 dollars, at a premium of 39 


fats on a hundred dollars. How much did the pre- 
nium amount to? 


| XV. 
INTEREST. 


_INTEREsT is a premium paid for the use of money. 
' Tt is computed by percentage; a certain per cent. on 
he money being paid for its use, for a stated time. 
) The money on which interest is paid, is called the 
rincipal. The per cent. paid, is called the Rate. The 
incipal and interest added together, are called the 
Imount. 
When a rate per cent. is stated without the mention of 
ny term of time, the time is understood to be 1 year. 
The rate of interest is regulated by state laws, and is 
ot uniform in all the states. We shall, however, first 
reat of 6 per cent. per annum, as this is the rate most 
ommonly paid. 
. 


| As interest is always expressed by some rate per cent., 
he most convenient way of computing it is, to find the 
‘eeimal expression of the rate for the time, and multiply 
fe principal by this decimal: the product is the interest. 
hus, if the rate for 1 year be 6 per cent. or .06, for 2 
fears itis 12 per cent. or .12, for 3 years it 1s 18 per 
ent. or .18, and soon. The interest of 24 dollars for 
} years, at 6 per cent. a year, is found thus, 24.18 = 
$4.32, the interest sought. 


| 


| 


94 ; - ARITHMETIC. x 


Rate PER CENT. FoR Montus. The decimal ey 
pression of the rate for months, when the rate is 6 pt 
cent. a year, is easily obtained; for, if the rate fore 1 
months be 6 per cents or .06, for 1 month it is 7 of 
per cent. which is 4 per cent. or .005; for 2 months 
is 1 per cent. or .01; for 3 months it is 15 per cent. ¢ 
.015; for 4 months it is 2 per cent. or 025 for 5 mont 
it is Qi per cent. or .025; for a year and L month it | 
64 per cent. or .065; for a year and 2 months it is 7 pe 
oan or .07; for a year and 11 months it is 114 per cen’ 

“115. i 
ve If the rate of interest be 6 per cent. for a year 


what is the rate for 1 month?...... for 6 months ?..... 
for 7 months ?....... for 8 months ?...... for 9 months? © 

2. At 6 per cent. a year, what is the rate for a yea 
and 1 month?...... a year and 3 months?...... a yea 
and 4 months ?...... a year and 10 months ? 


"i 
RATE PER CENT. FoR Days. Observe, that the rat 
for 2 specie which is 60 days, 1 is 1 per pained or .OL 
and for 75 of 60 days, which is 6 days, it is 7’ of 01 
which is .001. Now since the rate for 6 days is 1-thou 
sandth, the rate for any number of days is as many thou 
sandths as there are times 6 days. Therefore, to fin 
the rate for days, at 6 per cent. per annum, adopt thi 
following RULE. Denote the days as so many thou 
-sandths, and dwide the expression by 6: the quotien 
ay be the rate. | 
If the rate of interest be 6 Pe cent. for a year, wha 
is eh rate for 1 day?...... for 2 days?...... for 3 days , 
note for 4 days?...... for 5 days?...... for 6 days? ...1) 
for 7 days?...... for 9 days?...... for 24 days?...... foi 
26 days? 
4, At 6 per cent. a year, what is the rate for 2 months 
and 12 days?...... 3 months and 10 days?...... for 6 
months and 18 days?...... for 10 months and 29 days? 


5. What is the interest, and what the amount of 546 
dollars 72 cents, for 4 years 7 months 19 days, at 6 Ber 
cent. a year? 


» “4 
,} 


“XY. INTEREST. 95 
4 To find the rate for 4 years, we multi- 546.72 
ply the rate for 1 year by 4; thus, .06 X 27816 
-4=.24. To find the rate Fer 7 months, 398032 
we multiply the rate for 1 month by 7; 54672 


thus, .005 X 7=.035. To find the rate 437376 
jor 19 days, we denote 19 as eh alt dk 389704 
and divide the expression by 6; thus, 499344 
| 019+6—.00316-+. Now the sum ———-— 
of these rates, .24-++-.035-+ .00316—= 1 
: 27816, is the rate for the whole time; “*"'‘“ 
and by this sum we multiply the principal. 698.7956352 

‘The interest found, is $152.07,5+3; 7° 
“which, added to the principal, gives the amount, $698 
—79,5-+-. The rate for 19 days is not exact, as the deci- 
nal does not terminate; it is, however, sufficiently near 
exactness. 

6. What is the interest of 148 dollars 92 cents, for 3 

years, at 6 per cent. per annum ? 
7. What is the interest of 57 dollars 10 cents, for 5 
years, at 6 per cent. a year? 
-. 8. What is the interest of 93 dollars 50 cents, for 4 
rears, at 6 per cent. a year? 
9. What is the interest of 608 dollars 62 cents, for a 
year and 9 months, at 6 per cent. a year? 
| 10. What will 713 dollars 33 cents amount to, in 2 
rears and 10 months, at 6 per cent. per annum ? 

11. What will 1256 dollars 81 cents amount to, in 8 
nonths, at the rate of 6 per cent. a year? 
12. What is the interest of 100 dollars, for 1 year 1i 
nonths and 24 days, at 6 per cent. a year? 
13. To what sum will 37 dollars 50 cents amount, in 
| year 7 months and 21 days, at 6 per cent. per annum ? 

14. What is the interest of 314 dollars 36 cents, for 1 
year 1 month and 6 days, at 6 per cent. a year? 

15. What is the interest of 37 dollars a cents, for 
11 months and 15 days, at 6 per cent. a year f 

16. What is the interest of 512 dollars 38 cents, for 

7 months and 10 days, at 6 per cent. a year? 

17. To what sum will 691 dollars 28 cents amount, in 
1 year and 1 month, at 6 per cent. a year? 


96 ARITHMETIC, J 


18. What is the amount of 194 dollars 69 denies for 
year 5 months and 6 days, at 6 per cent. a year? 
19. What will 32 dollars 47 cents amount to, in| 
months and 25 days, at 6 per cent. a year? | 
20. What is the interest of 217 dollars 19 cents, fa 
year and 17 days, at 6 per cent. a year? 
21. What is the amount of 143 dollars 37 cents, fie 
year 9 months and 4 days, at 6 per cent. per annum ?- 
22. To what sum will 203 dollars 9 cents amount,” 
2 years and 18 days, at 6 per cent. per annum ? 3 
23. To what sum will 18 dollars 63 cents amount, 
1 year 10 months and 19 days, at 6 per cent. ayear 2” 
24. What is the interest of 600 dollars, for 7 mont! 
and 22 days, at 6 per cent. a year ? 
25. What is the interest of 817 dollars 44 cents, f 
11 months and 12 days, . at 6 per cent. a year? i 
26. What is the interest of 155 dollars, for 1 year 
sa and 10 days, at 6 per cent. a year? 
To what sum will 109 dollars 12 cents amount, | 
5 cab and § days, at 6 per cent. a year ? ‘a 
28. What is the amount of 25 dollars 92 cents, | 
_ year 4 months and 7 days, at 6 per cent. a year? 
29. To what sum will 65 dollars 48 cents amount, 
1 year 1 month and 18 days, at 6 per cent. a year? — 
30. What is the interest of 110 dollars 25 cents, fi 
10 months and 4 days, at 6 per cent. a year? 
» 31. What is the mterest of 2814 dollars jn cents, fi 
6 months and 3 days, at 6 per cent. a year? é 
32. What is the amount of 84 dollars 33 cents, for 
months and 26 days, at 6 per cent. per annum? 
33. What is the interest of 345 dollars 68 cents, fi 
7 months and 13 days, at 6 per cent. a year? : 
34. To what sum will 13 dollars 98 cents amount, , 
2 years 4 months and 7 days, at 6 per cent. a year? — 
35. What is the interest of 802 dollars 27 cents, for 
month and 5 days, at the rate of 6 per cent. a year! — 
36. What is the interest of 1309 dollars, ast 2 mont] 
and 3 days, at the rate of 6 per cent. a year? F 
37. To what sum will 23 dollars 8 cents amount, in! 
years 6 months and 22 days, at 6 per cent. a year? 


a 


7 


i XV. INTEREST. 97 


_* 38. What is the interest of 2538 dollars 17 cents, for 
_$months and 28 days, at the rate of 6 per cent. a year? 
' 39. What is the amount of 1800 dollars 34 cents, for 
1 year and 2 days, at 6 per cent. a year? 
40. What is the interest of 199 dollars 15 cents, for 1 
| year and 23 days, at 6 per cent. a year? 
_ 41. To what sum will 49 dollars 5 cents amount, in | 
year 2 months and 3 days, at 6 per cent. a year? 
_ 42. What is the interest of 201 dollars 50 cents, for 7 
| years, at 6 per cent? 
| 43. What is the interest of 3010 dollars 75 cents, for 
3 months and 1 day, at the rate of 6 per cent. a year? 
' 44. To what sum will 41 dollars 6 cents amount, in 1 
_ year 5 months and 14 days, at 6 per cent. a year? 
'* 45. What is the amount of 50 dollars and 11 cents, 
for 1 year and 21 days, at 6 per cent. a year ? 
46. What is the interest of 1100 dollars for a year 
and 15 days, at 6 per cent. a year? 
47. What is the interest of 9 dollars 89 cents, for 1 
year and 27 days, at 6 per cent. a year? 
48. What is the interest of 80 dollars, for 1 year 5 
__months and 12 days, at 6 per cent a year? 
_ 49. What is the*interest of 90 dollars, for 1 year 2 
_ months and 6 days, at 6 per cent. a year? 
50. Towhat sum will 55 dollars amount, in 3 years and 
9 days, at 6 per cent. a year? 
51. What is the amount of 4119 dollars 20 cents, for 1 
year and 5 days, at 6 per cent. a year? | . 


To compute interest by pays, when the rate 1s 6 per 
cent. per annum. RULE. Multiply the principal by the 
number of days, and divide the product by6. The quo- 
tient is the interest in mills, when the principal consists 
of dollars only; but when there are cents in the prinet- 
pal, cut off two figures from the right of the quotient, and . 
the remaining figures will express the mills. 

This rule—like the rule for finding the per cent. for 
days— is based upon the supposition of 360 days to the 
year; and, since the year contains 365 days, the rule 
gives 7}; part more than a true six per cent. terest. 

9 


yg ARITHMETIC. XWe 


| 
52. What is the interest of 86 dollars, for 20 days, 4 
6 per cent. a year? 
53. What is the amount of 108 dollars, for 25 days, 
at 6 per cent. a year? 
54. What is the interest of 204 dollars, for 40 days, at 
6 per cent. a year? 
55. What is the interest of 1000 dollars, for 29 dayal 
at 6 per cent. a year? | 


56. What is the amount of 98 dollars 60 cents, for 35) 


days, at 6 per cent. a year? 


57. What is the interest of 250 dollars, for 18 aayall 


at 6 per cent. a year? 


58. What is the interest of 61 dollars 25 cents, for 28) 


days, at 6 per cent. a year? 
59. What is the amount of 215 dollars 78 cents, for 50° 
days, at 6 per cent. a year? | 
60. What is ae interest of 71 dollars, for 41 days, at 
6 per cent. a year? 
61. What is the interest of 3333 dollars, for 10 days, | 
at 6 per cent. a year? 


62. What is the amount of 37 dollars 58 cents, for 16 


days, at 6 per ‘cent. a year? 

63. What is the interest of 91 doHars 80 cents, for 57. | 
days, at 6 per cent. a year? 

64. What is the interest of 4109 dollars, for 18 dayss | { 
at 6 per cent. ayear? 

65. What is the amount of 5214 dollars, for 50 days, 
at 6 per cent. a year? | 

66. What is the difference between the interest of | 
$1000 for 1 year, computed by the year, and the interest | 
on the same sum for the same time, computed by days; 
both at 6 per cent.? | 


It will be observed, that, in all the preceding examples, | 


the rate of interest has been 6 per cent. per annum. 


. The method of computing interest at any other rate per 


cent. is the same, and equally simple, when the time con- 
sists of years only; but when there are months and days inv 
the time, and the rate per cent. perannum is other than 6, - 
it will frequently be convenient to find the interest for a 


i ov. INTEREST. 99 


year first; and then for the months, to take the aliquot 


parts of a year; and for the days, the aliquot parts of a 


month; as in the following examples. 


67. What is the interest of 934 dollars 34 cents, for 3 


years and 5 months, at 7 per cent. per annum? 


934.34 
.07 


months is } of a Y. 3) 65.4038 interest for 1 year 


3 
196.2114 interest for 3 years. 


“Lmonth is fof 4 ms. 4) 21.8012 interest for 4ms. 


5.4503 interest for 1m. 
$§ 223.4629 interest for 3Y.5ms. 


68. What is the interest of 371 dollars 42 cents, for 
I year 9 months and 19 days, at 75 per cent. a year ? 
371.52 
.O75 
185760 «. 
260064 
6 months is 4 of a year. 2) 27.86400 for 1 year. 
3 months is} of 6ms. 2) 13.93200 for 6 months. 
15 days is ¢ of 3ms. 6) 6.96600 for 3 months. 
3 days is tof 15 days. 5) 1.16100 for 15 days. 
IdayisZof3days 3) .23220 for 3 days. 
.07740 for 1 day. 
$50.53268 for the whole time 


69. What is the interest of 412 dollars 17 cents, for 1 
year 7 months and 10 days, at 7 per cent. a year? 

70. What is the interest of 15748 dollars, for a year, 
at 44 per cent.? 

71. What is the interest of 125 dollars 50 cents, for 2 
years at 7 per cent. a year? , 

72. What is the interest of 969 dollars, for 4 years, 
at 8 per cent. a year? 

73. What is the interest of 655 dollars 30 cents, for a 
year, at 7 per cent.? 


| 
100 aR ee XV. f 


74. Whats is the interest of 404 cli 39 cents, for a ; 
ge at 54 per cent.? | 
aa i) ‘what sum will 1060 dollars 90 cents amount, ia 
a a at 7 per cent.? 
76. What is the interest of 1650 dollars, for a eu at 
30 per cent. | 
77. What will 1428 dollars amount to, in a year and 5 | 
months, at 5 per cent. a year? 
78. What is the interest of 2194 dollars 50 cents, for | 
a year and 10 months, at 7 per cent. a year? 
79. What is the interest of 20750 dollars 42 cents, | 
for 1 year 2 months and 20 days, at 44 percent. a year f 
80. What is the interest of 1109 dollars 44 cents, for | 
11 months, at 55 per cent. a year? 
81. What is the interest of 717 dollars 19 cents, foe 
5 months and 6 days, at the rate of 7 per cent. a year? 
82. What is the interest of 2119 dollars 78 cents, | 
for 3 months and 24 days, at 44 per cent. a year? | 
83. To what sum will 107 dollars 29 cents amount, in | 
7 months and 5 days, at the rate of 7 per cent. a year? | 
84. To what sum will 5128 dollars 60 cents amount, | 
in 3 months and 26 days, at 545 per cent. a year? } 
85. What is the mterest of 8244 dollars, for 1 month '. 
and 20 days, at the rate of 8 per cent. per annum? | 
86. What is the interest of 1062 dollars 80 cents, for 
2 months, at the rate of 9 per cent. per annum : P 
87. What is the interest of 4008 dollars 90 cents, for 
9 months, at the rate of 75 percent. a year? i 
88. What is the interest of 12416 dollars 25 cents, for | 
as pare at the rate of 4 per cent. a year? 
. To what sum will 103 dollars 70 cents amount, | 
in : year. 2 months and 13 days, at 7 per cent. a year? 
90. ‘Lo what sum will 86 dollars 21 cents amount, in | 
1 year 1 month and 27 days, at 7 per cent. a year? 
91. What is the interest of 502 dollars 9 cents, for 1_ 
year 3 months and 7 days, at 7 per cent. a year? 
92. What is the interest of 319 dollars 27 cents, for 
2 years 7 months and 11 days, at 7 per cent. a year? 
93. What is the amount of 753 dollars 50 cents, for 
} year 9 months and 21 days, at 30 per cent. a year? | 


@ 
“ei 


XV. INTEREST. 101 


94. To what sum will 207 dollars 8 cents amount, ir 
1 year 4 months and 5 days, at 7 per cent. a year? 
~95. What is the interest of 99 dollars 10 cents, for 2 
years 1 month and 23 days, at 7 per cent. per annum ? 


To calculate interest on English money, first reduce 
the shillmgs, pence and farthings, to the decimal of a 


_ pound; the operation will then be as simple as the opera- 
tion on Federal money. 


96. What is the interest of £17 10s. 6d. for 2 years 


_ 6 months, at 4 per cent. a year? 
£ : £ 


Sa £ £ 
17 10 617.525. Then, 17.525 *.10=1.7525. 
. £ SS ds. ari 
i 1.7625=1 15 0 2345 Ans. 
97. What is the interest of £42 18s. 9d., for 1 year 
7 months and 15 days, at 5 per cent. per annum? 
_ 98. What is the interest of £23 8s. 9d., for 6 years, 
at 7 per cent. a year? 
99. To what sum will £140 12s. 34d. amount, in 1 
year 4 months and 12 days, at 6 per cent. a year ? 
100. To what sum will £463 19s. 6d. amount, in 2 
years and 8 months, at 6 per cent. per annum ? 
101. What is the interest of £104 16s. 105d., for 


11 months and 27 days, at the rate of 7 per cent. a year ? 


102. What is the interest of £90 5s. 3d., for 1 year 
1 month and 9 days, at 7 per cent. per annum ° 
103. What is the interest of £512 7s. 4d., for I 


year 2 months and 21 days, at 5 per cent. a year? 


104. To what sum will £210 10s. 6d. amount, in 
1 year 3 months and 18 days, at 7 per cent. a year? 

105. What is the interest of £2148 13s. 3d., for 5 
months and 17 days, at the rate of 54 per cent. a year? 

106. What is the interest of £750 4s. 6d., for 2 
years 3 months and 20 days, at 7 per cent. ayear ? 

107. To what sum will £70 10s. amount, in 3 years 


2 months and 10 days, at 74 per cent. a year? 


108. What is the interest of £803 5s. 7d., for 10 
months and 14 days, at the rate of 5 per cent. per 
Q* 


annum ? 


102 ARITHMETIC. XV. 


109. To what sum will £13 13s. 6d. amount, in L 
year 11 months and 19 days, at 5 per cent. per annum. 


PARTIAL PAYMENTS. 


Tn computing interest on notes, bonds, é&c. whereon 
partial payments have been made, it is customary, when 
settlement is made in a year, or in less than a year from. 
the commencement of interest, to find the amount of the 
whole principal to the time of settlement, and also the | 
amount of each payment, and deduct the amount of all the 
payments from the amount of the: principal. | 

The learner may compute the interest on the follow 
notes; considering the rate to be 6 per cent. per annum, | 
when no other rate is stated. 


(PLOT Mae Boston, January 14th. 1833. 7 
For value received, I promise Samuel Burbank Jr. to | 
pay him or order the sum of one hundred and forty-one _ 
dollars and eight cents, in three months, with interest | 
afterward. Horace Chase. © | 
On the back of this note were the following endorsements. May | 
Ist. 1833, received seventy-five dollars. September 14th. 1833, re- ) 


ceived forty-five dollars. The balance of the note was paid January | 
14th. 1834. How much was the balance ? *| 
First payment, $75. | 2nd. payt. $45. | Principal, $ 141.08 ) 
Interest,8m.14d.,  3.17/Int., 4m. 90 | Int. 9m. 6.34 | 
Amount, $ 78.17 Amount, 45, 90 Amount, 447.42 it 

8.17 124.07 | 


Amount of payments, $ rie Balance, $23.35 | 


| 
| 


(111.) '  - New York, May 25th. 1833. 
For value received, I promise Joseph Day to pay him) 
or order the sum of three hundred and oné dollars and | 
forty-seven cents, on demand, with interest. j 
Attest. John Smith. Samuel Frink. 
On the back of this note, the following endorsements were “made, 
July Ist. 1833, received sixty-seven dollars and fifty cents. Janu- | 
ary 4th. 1834, ‘received forty-eight dollars. April ‘11th. 1834, re- 
by ‘ived thirty-nine dollars. "The balance of this note was paid June 
ist. 1884. Required the balance. 


Xv. INTEREST. 103 


(112.) ‘Philadelphia, June 26th. 1833. 
For value received, I promise Charles 8. Johnson to 
pay him or order ninety-three dollars and twenty-eight 
cents, on demand, with interest. James Orne. 
Attest. Levi Dow. 


_ On this note there were two endorsements, viz. Nov. 5th. 1833, 
received forty-three dollars and seventy-five cents. Feb. 22d. 1834, 
‘received thirty-seven dollars. What was due, May 26th. 1834, 
when the balance was paid. 


(113.) Baltimore, March 4th. 1832. 
For value received, I promise Hay & Atkins to pay 
them or order the sum of four hundred and three 
dollars and fifty-six cents, in nme months, with interest 
afterward. Homer Chase. 


The following endorsements were made on the back of this note. 
Jan. ist. 1833, received one hundred and eighty-four dollars. 
August 18th. 1833, received one hundred dollars. ‘This note was 
taken up Dec. Ist. 1833. What was the balance then due upon it? 


| 


(114.) Hartford, July 11th. 1831. 

For value received we promise Joseph Seaver to pay 
him or- order the sum of two hundred and seventeen 
dollars and fifty cents, in four months, with interest after 
that time. Whiting & Davis: 

On this note there were three endorsements: viz. Nov. 16th, 
1831, received ninety-three dollars. Feb. 12th. 1832, received 
fifty dollars. August 2d. 1832, received sixty-seven dollars and 
seventy-five cents. This note was taken up Oct. 4th. 1832. How 
much was then due upon it? 


(115.) Burlington, October Ist. 1832. 
For value received, we promise Hannum, Osgood, & 
Co. to pay them or order the sum of seven hundred and 
fourteen dollars, in three months, with interest afterward. 
Mason & Gould. 
The following payments were endorsed on the note. January 
| Ist. 1833, received three hundred and sixty-four dollars. May Ist. 
_ 1833, received one hundred and twenty-five dollars and fifty cents. 
August Ist. 1833, reccived eighty-six dollars. Nov. Ist. 1833, re- 
| ceived a hundred and ten dollars. ‘The balance due on this note 
| was paid Jan. Ist. 1834. How much was it? 


| 


| 


104 ARITHMETIC. XV. 


If settlement is not made, till more than a year has_ 
elapsed after the commencement of interest, the preced-_ 
ing mode of computing interest, when partial payments 
have been made, ought not to be adopted; and indeed de 
is not in strict conformity with law. i 

The United States Court, and the Courts of the several 
States, in which decisions have been made and reported, 
with the exception of Connecticut and Vermont, and 
a slight variation in New Jersey, have established a gen-_ 
eral rule for the computation of interest, when partial 
payments have been made. This rule is well expressed 
in the New York Chancery Reports, in a case decided by — 
chancellor Kent, and here given in the Chancellor’s own) 
words, as follows. 4 

‘¢ The rule for casting interest, when partial payments 
have been made, is to apply the payment, in the first placa 
; 
" 


; 
to the discharge of the vnterest then due. If the payment 
exceeds the interest, the surplus goes towards discharging 
the principal, and the subsequent interest rs to be com-| 
puted on the balance of principal remaining due. bs | 
payment be less than the interest, the surplus of interest) 
must not be taken to augment the principal; but interest 
continues on the former principal until the period when 
the payments, taken together, exceed the interest due, 
and then the surpius is to be applied towards discharging) 
the principal; and interest is to be computed on the bale 
ance, as aforesaid.’ i 

The interest on the following notes, must be computed | 
by the above legal rule. | | 


iE 


i 


(116.) Washington, March 4th. 1832. 
For value received, I promise Nehemiah Adams to 
pay him or order the sum of one thousand two hundred 
dollars, on demand, with interest. Charles Train. — 
Attest. William Dorr. 


The following endorsements were made on this note. June 
10th. 1832, received one hundred and sixty-nine dollars and twenty | 
cents. Oct. 22d. 1832, received twenty dollars. March 30th. 1833, 
received twenty-eight dollars. Nov. 5th. 1833, received six hun- 
dred and eighteen dollars and five cents. What was the balance | 
due, on taking up this note, March 5th. 1834? . 


( 
| 
| 
| 
| 
| 


3.4%. INTEREST. 105 


7 Principal, - $ 1200. 
_ Interest from Mar. 4, to June 10, (3 m. 6 d.), - = 19.20 
First Amount, - - 1219.20 
. First payment - - : - - - - 169.20. 
Balance, forming a new principal, . - - - - 1050,00 
Interest from June 10, to Oct. 22,(4m.12d.), $23.10 — 
| Second payment, - - - - . 20. 

3.10 


| Leaving interest unpaid, - - -~ - 
Interest from Oct. 22, to Mar. 30, (5m. 8 d.), 27.65 


Third payment, Sie ae ak ta 28.00 
‘Leaving interest unpaid, = - - - - 2.75 
Interest from Mar. 30, to Nov. 5,(7m.6d.), 37.80 40.55 


| Second Amount, - - 1090.55 
Fourth payment, - - - - - - - 618,05 
Balance, forming a new principal, - - - - 472.50 
Taterest from Nov. 5, to Mar. 5, (4 m.), - - - 9.45 
Balance due on taking up the note, - - - - $481.95 
BElv)) Richmond, Jan. 5th. 1833. 


For value received, I promise Joseph Tufts to pay 
him or order one hundred and forty-three dollars and 
fifty cents, on demand, with interest. John Hanes. 

Two payments were endorsed upon this note: viz. April 13th. 
1833, received forty-five dollars and eighty-four cents. Dec. 22d. 


1833, received fifty-four dollars and fifteen cents. The balance of 
this note was paid March 28th. 1834. How much was it? 


(118.) Raleigh, July Ist. 1832. 

For value received, I promise Charles Goodrich to 
pay him or order the sum of six hundred and twenty-five 
dollars and fifty cents, in three months, with interest after- 
ward. John Frink. 

Three payments were endorsed upon this note: viz. January 
Ist. 1833, received two hundred dollars. Nov. Ist. 1833, received 


twenty dollars. Jan. ist. 1834, received three hundred dollars. 
, The balance was paid May Ist. 1834. How much was it? 


(119.) Charleston, Dec. 22d 1830. 
_ For value received, I promised George Winship to 
pay him or order ninety-seven dollars and eighty cents, 
on demand, with interest, Thomas White. 


, 
| 


| 
rs ' 
I 

I 


106 ARITHMETIC. | XV. 


The endorsements made on this note were the following. Oci. 
12th. 1831, received twelve dollars eighty-five cents. July 20th 
1832, received twelve dollars and seventeen cents. Feb. 26th 
1833, received fourteen dollars and ninety-five cents. August 26th 
1833, received thirty-six dollars and ten cents. Requir ed the | 
ance, which was paid Jan. 31st. 1834. 


} 
| 
| 


(120.) Augusta, January Ist. 1331. _ 
For value received, I promise Israel Capen to pay 
him or order eighty- -four dollars and forty cents, on de_ 
mand, with interest. Edward Ruggles. | 
On the back of this note were the following endorsements. Oct 
9th. 1831, received nineteen dollars and thirty-two cents. Juh 
15th. 1832, received twenty dollars. April 9th. 1833, receiver 
twenty-one dollars and eighty-one cents. Oct. 9th. 1833, receive( 


twenty-two dollars and fifteen cents. The balance of this not. 
was paid Feb. 19th. 1834. How much was it? ) 
t 


(121) New Orleans, Feb. 22d. 1830. 
For value received, I promise Maynard and Noye: 
to pay them or order the sum of nine hundred dollars, i ir 
three months, with interest till paid. Isaac Jettison. 
Attest. William Proctor. 


The following payments were endorsed upon the note. May 
22d. 1830, received twenty-five dollars. Sept. 22d. 1830, received; 
fifteen dollarse May 22d. 1831, received thirty-five dollars. May 
22d. 1832, received one hundr ed and forty-five dollars and twelve 
cents. Dec. 4th. 1832, received one hundred and twenty-five 
dollars and sixty cents. "May 22d. 1833, received two hundred and’ 
nineteen dollars and sixty cents. Dec. 31st. 1833, received two! 
hundred and sixty-eight dollars and twenty-five cents. The. 
Si ist Ae this note was paid Feb. 24th. 1834. What was the 

alance: 


(122.) Cincinnati, Dec. Ist. 1830. 

For value received, I promise Horatio Davis to pay 
him or order the sum of one thousand dollars, on demand, 
with interest tll paid. Edward.Lang. | 


Five partial payments were endorsed on this note: viz. Feb.: 
Ist. 1832, received seventy-five dollars. June Ist. 1832, received’ 
twenty dollars. August Ist. 1833, received twenty dollars. October 
Ist. 1833, received seven hundred and fifty dollars. Feb. 1st. 1834, 
received one hundr ed dollars. ‘The balance of this note was paid 

- J une Ist. 1834, How much was it? 


~XV.~ INTEREST. 167 


(123.) Louisville, April 4th. 1832. 
_ For value received, I promise Samuel H. Wheeler to 
‘pay him or order the sum of three hundred and ninety- 
six dollars, on demand, with interest, at the rate of 7 per 
cent. a year, till paid. George Guelph. 

Partial payments were made on this note, as follows: Sept. 14th. 
1882, received twelve dollars. May 4th. 1833, received eighteen 
dollars. Oct. 24th 1833, received forty-nine dollars twelve cents. 
The balance was paid May 30th. 1834. What was the balance ? 


(124.) Nashville, Sept. 7th. 1831. 
For value received, I promise Darius Pond to pay 
him or order the sum of four hundred and eighty-six 
dollars and ninety cents, on demand, with interest at the 
irate of 7 per cent. a year. Martin Smith. 

The following partial payments were endorsed on this note. 
March 22d. 1832, received one hundred and twenty-five dollars. 

Nov. 29th. 1832, received one hundred and fifty dollars. May 
‘A8th. 1833, received one hundred and twenty dollars. ‘The bal- 
ance was paid April 19th. 1834. Required the balance. 


(125.) Albany, August 13th. 1830. 
For value received, I promise Theodore Leonard to 
pay him or order the sum of two hundred and ninety- 
eight dollars and nineteen cents, on demand, with interest 
at the rate of 7 per cent. a year. Stephen Kirkland. 
Attest. W. Stevenson. 


The following endorsements were made on this note. April 
6th. 1831, received fifty-four dollars. Dec. 17th. 1831, received 
forty-two dollars. June, 21st. 1832 received sixty-one dollars. 
Feb. 26th. 1833, received thirty-seven dollars and eighty cents. 
July 8th. 1833, received seventy-five dollars. The balance was 
paid May 12th. 1834. How much was the balance ? 


COMPOUND INTEREST. 


Compound interest is that which is paid not only for 
the use of the principal, but also, for the use of the inter- 
est after it becomes due. 

_ When the interest is payable annually, find the interest 
for the first year, and add it to the principal, and this 
amount is the principal for the second year. Find the 


108 


interest on this second principal, and add as before; thi! 
amount is the principal for the third year: and so o7 
through the whole number of years. When the interes| 
is payable half-yearly, or quarterly, find’ the interest fo! 
half a year, or a quarter of a year, and add it to thi 
principal, aud thus proceed through the whole time} 
Subtract the first principal from the last amount, and thi 
remainder is the compound interest. | 

126. What is the compound interest of a thousam 
dollars for 3 years, at 6 per cent. per annum ? 


¢ 1000. 
60. 


1060. 
63.60 


1123.60 
ae ee 416 


———— 


1191.016 
1000. 


$191.016 


127. What is the compound interest of 740 dolla fo 
6 years, at 6 per cent. per annum ? 

128. What is the compound interest of 500 dollar 
for 4 years, at 7 per cent. per annum? | 

129. To what sum will 450 dollars amount, in 5 years 
at 5 per cent. per annum, compound interest ? | 

130. What is the compound interest of £760 10: 
for 4 years, at 4 per cent. per annum? 

131. A gave B a note for 300 dollars, with interest ; 


ARITHMETIC. XV 


principal. 
interest for the first year. 


amount, principal for the second year. | 
interest for the second year. 


second amount, principal for third year 
interest for the < _ird year. 


third amount. | 
first principal deducted. 


Answer. 


— 


6 per cent. a year, payable semiannually. How mue i 


did it amount to in 2 years, at compound interest ? 
132. At compound interest, what will 600 dollay 

" amount to in 14 year, at the rate of 6 per cent. a yea 
interest payable quarterly ? a 


PROBLEMS IN INTEREST. a 


' In reviewing the subject of simple terest, we percely 
four several problems, which arise from its conditions 
and which we shall now distinctly notice. 


| 


XV. INTEREST. 109° 


PROBLEM I. The principal, time, and rate per cent. 
given, to find the interest. 

RULE. Multiply together the decimal expressing the 
rate per annum, the lime in years and the decimal 1 of a 
year, and the principal: the product will be the interest. 

This problem has already been exemplified in the pre- 
ceding pages of this article. 


PROBLEM II. The principal, time, and amount given, 
to find the rate per cent. per annum. 
RULE. Subtract the principal from the amount, and 


the remainder will be the interest for the given time. 


4 
4 


Divide this interest by the given time expressed in years 
or the decimal of a year, and the quotient will be the 
‘interest for one year. Divide the interest for one year by 
‘the given principal, and the quotient will be the rate per 
cent. per annum. 

133. At what rate per cent. per annum must 172 dol- 
lars 40 cents be put on interest, in order to amount to 
332 dollars 74 cents, in 5 years : ? 

134. Lent 51 dollars 25 cents, and in 1 year and 4 
months it amounted to 55 dollars 35 cents. What wa 
the rate per cent. per annum? 

135. Borrowed 340 dollars for 9 eertnes and at the 
expiration of the time it amounted to 355 dollars 30 cents. 


| What was the rate of interest per annum? 


| 


| 


136. At what rate per cent. per annum must 874 cents 


‘be put on interest, in order to amount to 98 cents, in2 years ? 


PROBLEM Ill. The principal, rate per cent., and 
amount given, to find the time. 
RULE. Subtract the principal from the amount, and 


. the remainder will be the interest. Divide the interest oy 


the principal, and the quotient will be the interest of 1 
dollar. Divide the interest of 1 dollar by the rate, and 
the quotient will be the time. 

137. In what time will 89 dollars 25 cents amount to 
92 dollars 82 cents, at the rate of 6 per cent. a year? 

138. In what time will 171 dollars 40-cents, amount to 
231 dollars 39 oe at 7 per cent. a year? 

0 


110 ARITHMETIC. XME! 


139. Borrowed 163 dollars 50 cents at 6 per cent. a 
year; at the time of payment it amounted to 176 dollars 
58 cents. How long did I keep the money ? 

140. In what time will 4810 dollars 25 cents, amount 
to 5002 dollars 66 cents, at 6 per cent. a year? 

141. Lent 114 dollars at an interest of 7 per cent. a) 
year; on its return it amounted to 127 dollars 30 cents. — 
How long was it out? 

142. In what time will $100, or, any other sum of k 
money double, at the rate of 6 per cent. per annum, ! 
simple interest ? 


. as Soe 


PROBLEM IV. The amount, time, and rate per cent. 
given, to find the principal. 

RULE. . Divide the amount by the amount of 1 dollar i 
for the time, and the quotient will be the principal. | 

This problem-forms the subject of the next article, | 
under the head of Discount. 


XVI. | 
DISCOUNT 


Discount is an allowance made for the payment of | 
money before it is due. | 

The present worth of a debt, payable at a future period 
without interest, is that sum of money, which, being put 
on interest, arid amount to the debt, at the period when 
the debt is payable. 

It is obvious, that, when money is worth 6 per cent. 
per annum, the present worth of $1.06, payable in a_ 
year, is $1. Hence, the present worth of any debt, pay> 
able in a year, is as many dollars as there are times $1.06 _ 
in the debt. And hence we deduce the following. | 

RULE. Divide the debt by the amount of 1 dollar for i 
the time, and the quotient is the present worth. Subtract | 
the present worth from the debt, and the remainder will 
be the discount. 


wen mM. Oo 


XVI. DISCOUNT. | 111 


1. What is the present worth of 450 dollars, payable 
in 6 months, when money is worth 6 per cent. per 
annum ? 

2. What is the present worth of 535 dollars, payable 
in 15 months, when money is worth 6 per cent. per 
annum? 

3. When money is let for 6 per cent. per annum, 
what is the present worth of a note for 1530 dollars, 
payable in 18 months? 

4. Sold goods to the amount of 1500 dollars, to be 
paid one half in 9 months, and the other half in 18 months: | 
what is the present worth of the goods, allowing interest 
to be 5 per cent. per annum? 

5. What is the present value of a note for 2576 dol- 
lars and 83 cents, payable in 9 months, when interest is 
6 per cent. per annum? 

6. When interest is 6 per cent. a year, what is the 
difference between the discount on 1285 dollars for a 
year and 8 months, and the interest of the same sum for 
the same time ? 

7. Purchased goods amounting to 6568 dollars 50 
cents on a credit of 8 months: allowing money to be 
worth 4 per cent. a year, how much cash down will pay 
the bill ? 

8. A man, having a horse for sale, was offered for it 
225 dollars, cash in hand, or 230 dollars payable in 9 
months: he chose the latter, although money was worth 
7 per cent. a year. How much did he lose by his 
ignorance ? 
~ 9. Bought a quantity of goods for 1831 dollars 53 
cents cash, and the same day sold them for 1985 dollars 
48 cents on a credit of 6 months, when money was 5 
per cent. a year. How much did I gain upon the goods ? 

10. What is the discount on 198 dollars 60 cents, for 
9 months, when interest is 5 per cent. a year? 
~ 11. What is the discount on 241 dollars 81 cents, for 
7 months, when interest is 44 per cent. a year ? 

12. What is the present worth of 741 dollars 65 cents, 
payable in 48 days; interest being 6 per cent.? 


112 ARITHMETIC. XVII. 


XVII. 
BANKING. 


A BANK is an institution which trafficks in money. It 
is owned in shares, by a company of individuals, called _ 


stockholders; and its operations are conducted by a Presi- 


dent and board of Directors. It has a deposite of specie, 


and issues notes or bills, which are used for a circulating 


medium, as money. ‘These bills are mostly obtained 


from the bank in loans, on which interest is paid; and the 


amount of bills issued being greater than the amount of 


specie kept in deposite, a profit accrues to the bank. 
The interest on money hired from a bank, is paid at 


the time when the money is taken out—the hirer receiving 


as much less than the sum he promises to pay, as would 


be equal to the interest of what he promisesdo pay, from 


the time of hirmg the money until the time it is to be 
paid. From this circumstance, the interest on money 
hired from a bank is called discount, and the promissory 
note received at the bank is said to be discounted. 

A note, to be discounted at a bank, is usually made 
payable to some person, who endorses it, and who there- 
by binds himself to pay the debt, in case the signer of 
the note should fail to do so. Any person, therefore, 


who holds the note of another, payable at a future time, 
may endorse it, and obtain the money for it at a bank, — 


by paying the bank discount; provided the credit of the 
parties is undoubted. 

It is customary in banks, to compute the discount on 
every note for 3 days more than the time stated in the 
note; and the debtor is not required to make payment 
until 3 days after the stated term of time has elapsed. 
These-3 days are called days of grace. 


1. What is the bank discount on 775 dollars for 30. 


days, and grace, when interest is 6 per cent. a year? 
2. What is the bank discount on 900 dollars for 90. 
days, and grace, at the rate of 6 per cent. a-year? 


XVII. BANKING. 1138 


3. How much is received on a note for 2540 dollars’ 
80 cents, payable in 4 months, discounted at a bank, 
when interest is 45 per cent. a year? 

4. A note for 452 dollars, payable in 7 months, is 
discounted at a bank, when interest is 6 per cent. per 
annum. What sum is received on it? 

5. A note for 3000 dollars, payable in 70 days, is dis- 

counted at a bank, when interest is 6 per cent. a year. 
What sum is received on it? : 
- 6. Amerchant bought 1625 barrels of flour for 5 dollars 
a barrel cash, and on the same day sold it for 5 dollars 
60 cents a barrel, on a credit of 8 months, took a note 
for the amount, and got it discounted at a bank, when 
money was 6 per cent. a year. How much did he gain 
on the flour ? 

7. A man got his note for ¢ 1000, payable in 3 months, 
discounted at a bank, at the rate of 6 per cent., and im- 
mediately put the money he received for his note on 
interest for 1 year, at 6 per cent. He kept the money 
from the bank 1 year, by renewing his note every 3 
months, and paying in the required bank discount at each 
renewal. At the end of the year he received the amount 
of the money he had put on interest, and paid his note at 
the bank. How much did he lose by this exchange ? 

In the above example, interest on the several discounts 
paid into the bank forms part of the loss. 

8. A money broker subscribed for 20 shares in a new 
bank; at $100 a share. When the bank commenced 
operation he paid in 50 per cent. of the price of his stock, 
and in 6 months after, he paid in the remainder. In 12 
months from the time the bank commenced, there was a 
dividend of 34 per cent. on the stock among the stock- 
holders; aud the same dividend accrued every 6 months 
thereafter. At the end of 3 years the broker sold his 
stock at 7 per cent. advance. Now, allowing that this 
broker hired his money, and paid 6 per cent. annually, 
how much did he make by the speculation ? 

In this example, the broker must charge annual interest 
on the interest he pays, and must give credit for annual 
interest on his share of the dividends. | 

10% 


f 
114 ARITHMETIC. XVI. 


XVII. | 
EQUATION OF PAYMENTS. 1 


EQuATION OF PAYMENTS consists in finding a mean 
time for the payment at once of several debts, payable at 
different times, so that no loss of interest shall be sustained” 
by either party. 

For instance, if A owes B one dollar, payable in 1 oe | 
months, another dollar payable in 3 months, and a third | 
dollar payable in 4 months, at what time may the three 
sums be paid at once, without injustice to either of them 
It is evident, that the interest of 1 dollar for 2 months, is— 
the same as the interest of 2 dollars for 1 month; and, the 
interest of 1 dollar for 3 months, is the same as the inter- 
est of 3 dollars for 1 month; and the interest of 1 dollar _ 
for 4 months, is the same as the interest of 4 dollars for 1 
month: 2 dollars, 3 dollars, and 4 dollars, added together, 
make 9 dollars for 1 month; but the three sums to be~ 
paid, when added together, make only 3 dollars, which 

sum being only a third part of 9 dollars, the term of | 
credit must be three times as long, or 3 months, which 
is the equated time. This result is obtained by multiply-— 
ing the sum, payable in 2 months, by 2; that payable in 
3 months, by 3; and that payable in 4 months, by 4; and 
then adding the several products together, and dividing 
the sum of them by the sum of the debts. 

RULE. JWMultiply each debt by the time, in which tt is 
payable, and divide the sum of the products by the sum 
of the debts: the quotient will be the equated time. ; 


If I owe you 50 dollars payable in 4 months, | 
ie payable im 6 months, and 100 dollars payable in’ 
7 months, in what time may the three sums be paid at 
once, without loss to either of us ? 

2: A owes B 200 dollars, a0 dollars of which is to be 
paid 1 in 3 months, 60 dollars in 5 months, and the remain- 
dor in 10 months. At what time may the whole be paid 
at once, without injustice to either party ? tr 


/ 


| XVII. EQUATION OF PAYMENTS. 115 


_ 3. Bought goods to the amount of 1552 dollars, pay- 
able at four different times, as follows; 225 dollars and 
75 cents in 4 months, 250 dollars and 25 cents in 6 
months, 425 dollars and 50 cents in 8 months, 650 dol- | 
lars 50 cents in 10 months; but afterward agreed with 
my creditor to pay him all at once, at the equated time. 
‘What was the time? 

_ 4. If lowe you three sums of money payable at differ- 
ent times, viz. 50 pounds in six months, 60 pounds in 7 
months, and 80 pounds in 10 months, what is the equated 
‘time for paying the whole at once? 

5. Bought goods to the amount of 1000 dollars, 200 
‘dollars of which was to be paid down, 400 dollars in 5 
months, and the remainder in 15 months; but it was 
afterward agreed, that the whole be paid at once. In what | 
time ought the payment to be made ? ; 

6. A merchant has due to him a certain sum of money, 
to be paid as follows; ¢ in 2 months, 4 in 3 months, and 
the rest in 6 months. What is the equated time for pay- 
ing the whole ° | 

7. Sold goods amounting to 1296 dollars, of which 346 
dollars was to be paid in 24 months, 323 dollars in 6 
months, and the balance in 10 months; but the purchaser 
afterward agreed to make but one payment of the whole. 
What term of credit ought he to have? 

8. Bought goods to the amount of 640 dollars 80 
cents, payable } down, + in 4 months, 4 in 8 months, 
and the balance in a year; but afterward made an agree- 
ment to pay the whole at one time. In what time ought 
Ito pay for the goods ° 

9. A merchant has due to him $3800 to be paid in 60 
days, $500 to be paid in 120 days, and $750 to be paid 
in 120 days. What is the equated time for these dues ? 

10. A owes B $1200, to be paid in 8 months; but A 
offers to pay $400 in 4 months, on condition that the 
remainder shall continue unpaid an adequate term of time. 
In what time ought the remainder to be paid ? 

11. If adebt of $1000 be payable at the end of 7 
months, and the debtor agree to pay $300 at present, 


ba is the proper time for paying the rest? 


| 


116 ARITHMETIC. XIX 


XIX. 
PROFIT AND LOSS. 


The ascertaining what is gained or lost n buying anq 
selling, and the adjusting of the price of goods so as tc 
gain or lose a certain sum, or a certain per cent., comé 
under the head of Profit and Loss. 


_ 1. Bought a piece of broadcloth containing 28 ya 
for 112 dollars, and sold it at 5 dollars 25 cents a yard) 
How much, and what per cent. was my profit? (See 
Arr. xiv, Example 45.) | 

ie Bought 3 pieces of broadcloth, containing 28 yards, 
each, at 5 dollars 25 cents a yard. At what price pet 
yard must I sell it, to gain 20 per cent.? 

3. Bought cloth at 4 dollars 60 cents a yard, which, 
not proving so good as I expected; I sold at 3 dollars 91 
conts a yard. What per cent. did I lose? 

4. Bought 1250 barrels of flour for 6250 dollars. Al 
what price per barrel must I sell it, to make a profit of 
124 per cent.? 

5. Bought 30 hogsheads of molasses, at 20 dollars ¢ 
hogshead, in Havana; paid duties 20 dollars 66 cents; 
freight 40 dollars 78 cents; porterage 6 dollars 5 cents; 
insurance 30 dollars 84 cents. What per cent. shall 1 
gain by selling at 26 dollars per hogshead ? 

6. Bought wheat at 75 cents a bushel; at what price 
per bushel must I sell it, to gain 20 per cent.? 

7. A merchant received from Lisbon 180 casks al au 
raisins, containing 80$1b. each, which cost him 2 dollars 
18 cents a cask. At what price per cwt. must he sell) 
them, to gain 25 per cent. ? 4 

8. If I sell sugar at 8 dollars per cwt., and thereby, 
lose 12 per cent., what per cent. do I gain or lose, by 
selling the same at 9 dollars per ewt.? 

9. If I purchase 6 pipes of wine for 816 dollars, aul 
sell it at 59 dollars 50 cents a hhd. do I gain or lose, and 
what per cent.? | 

| 
| 


‘Xx. PARTNERSHIP. Lie 


10. If you purchase 5cwt. Igr. 12Ib. of rice, at 2 
dollars 80 cents per cwt., at what price per pound must 
you sell it, to make 6 dollars on the whole ? 

i. If I purchase 13cwt. of coffee at 124 cents per 
pound, at what price per lb. must I sell it, “to a 80 
dollars 8 cents on the whole? 

12. A miller sold a quantity of corn at 1 dollar a 
bushel, and gained 20 per cent.; soon after, he sold of 
the same, to the amount of $37.50, and gained 50 per 
cent. How many bushels were there in the last paren, 
and at what did he sell it per bushel ? 


XX. 
PARTNERSHIP. 


 Partnersulip is the union of two or more individuals 
\ ° ° 

mtrade. ‘The company thus associated is called a firm: 
wind the amount of property, which each partner puts 
ato the firm, is called his stock in trade. The profit or 
‘oss is shared among the partners, when the stock of each 
s employed an equal length of time, in proportion to 
ach partner’s stock in trade; but, when the stock of 
he several partners is employed in the firm unequal 
‘erms of time, in proportion to each one’s stock and the 
ime it was employed. 


1. A, B, and C entered into partnership, and the stock 

of each was employed in the firm one year. <A put in 
240 dollars, B 360 dollars, and C 120 dollars. They 
sained 350 dollars. What was each partner’s share of the 
rain ? 
_ We find in this example, that the whole capital of the 
lirm was 720 dollars. A’s stock was 240 dollars, and 
ie must have 24° of the gain. B’s stock was 360 dollars, 
ind he must have #§° of the gain. C’s stock was 120 
iollars, and he must have 129 of the gain. Observe the 
ollowing statement. . 


| 


S| 
; 
’ 


118 ARITHMETIC. Xx 
e502; and 2 of $350. is 1162 dollars, A’s share 
#$0—=25 and 2of $350. is 175 dollars, B’s share 
7se7-¢3 and tof $350. is 584 dollars, C’s share 


$350 Proof. } 
2. Messrs. Ralph Wheeler, Samuel Slade, and Jame) 
Libbey formed a connexion in business under the firm 0, 
Wheeler, Slade, and Libbey. Wheeler put into the firn 
2500 dollars; Slade 2000 dollars; and Libbey 1500 doll 
lars. ‘The stock of the several partners was in trade th) 
same term of time, and they gained 1500 dollars. Wha 
was each partner’s share of the profit ? ; 9 
3. Messrs. Joel Haven, Israel Varnum, Tyler Penni 
man, and James Conant formed a partnership under thi 
firm of Haven, Varnum, and Go. Haven put into thi 
firm 4000 dollars, Varnum 2500 dollars, Penniman 150 
dollars, and Conant 750 dollars. They traded in partner, 
ship 3 years, and gained 1750 dollars. How much wai 
each partner’s share of gain? 4q 
4. A, B, C, and D traded together one year. A pu 
in 800 dollars, B 500 dollars, C 300 dollars, and D 15¢ 
dollars; but by misfortune they lost 350 dollars. Wha’ 
loss did each partner sustain ? 4 
5. A gentleman dying, left two sons and a daughter, 
to whom he bequeathed the following sums; viz. to the 
elder son 1200 dollars, to the younger, 1000 dollars, anc. 
to the daughter 800 dollars; but it was found that his| 
whole estate amounted only to 750 dollars. How muck) 
did each child receive from the estate ? 
6. Three merchants bought a ship, for which they 
gave 8000 dollars. A paid 2850 dollars, B 1980 dollars, 
aid © the rest: m her first voyage she cleared 6400 
dollars. How much of the profit had each partner ? 
7. A and B traded together. A put into the firm 540 
dollars, and B the rest: they gained 387° dollars, of, 
which B’s share was 225 dollars. What was ’A’s gain, 
and what was B’s stock ? | 
8. The capital stock in the firm of Farmer, Turner, 
and Hancock, was 18477 dollars 60 cents. Farmer’s, 
stock was 9238 dollars 80 cents; Turner’s 6929 dollars 
10 cents; and Hancock’s the remainder. The stock 


; 
| 


XX. PARTNERSHIP. 119 


of the several partners was in the firm the same term of 
time: by misfortunes of various kinds, they lost 12375 
dollars 20 cents. What loss did each partner sustain ? 
9. <A, B, and C traded in partnership. <A’s stock 
was 385 ola 50 cents; B’s 297 dollars 75 cents; C’s 
175 dollars 25 cents: they gained 343 dollars 40 cents. 
erat v was each one’s share of gain? 
' When the stock of the several partners is in the firm 
unequal terms of time, the profit or loss must be appor- 
tioned with reference both to stock and time. Thus, 
A, B, and C, traded in company; A put in 200 dollars 
for 3 anal B 180 dollars for 5 months, and C 70 dol- 
lars for 10 months: they gained 132 dollars. Now, to 
apportion this gain justly, we say that A’s 200 dollars 
for 3 months was the same as 600 dollars for 1 montL; 
B’s 180 dollars for 5 months the same as 900 dollars for 
‘Tmonth; and C’s 70 dollars for 10 months the same as 
700 dollars for 1 month; therefore it is the same as if A 
nad put in 600 dollars, B 900 dollars, and C 700 dollars, 
ll for an equal term of time. These sums added together 
nake 2200 dollars; therefore, A had 33%% of the gain, B 
3200? and C 30°. These fractions, when reduced, are 
2) 23) ad 5. gy Of 192 dollars is 6 dollars; then A 
aad 6 times 6 dollars, B-9 times 6 dollars, and C 7 times 
3 dollars. 
RULE. Multiply each partner’s stock by the time tt 
vas in the firm; make each product the numerator of a 
raction, and the sum of the products a common denomi- 
vator; then multiply the whole gain or loss by each of 
hese fractions, for each partner’s share. 


10. A, B, and C traded in company. A putin 400 
tollars Dy 9 Coaonttid, B 300 dollars for 6 months, and C 
200 dollars for 5 months: they gained 320 dollars. 
What was the gain of each? 

11. X, Y, and Z formed a partnership. X put mto 
he firm 500 dollars for 18 months, Y 380 dollars for 13 
months, and Z 270 dollars for 9 months; but they lost 
RP dollars 50 cents. What was the loss of each ? 


120 ARITHMETIC. x3 


12. R and S entered into partnership for 16 month) 
R put in at first $1200, and at the end of 9 montl 
$200 more. S put in at first $1500, and after 6 montl| 
had elapsed he took out $500. In this partnership the 
gained $772.20. How must the gain be divided? | 

13. On the first day of January, A began business will 
380 dollars; on the first day of May following, he too 
B into partnership with 270 dollars; on the first da 
of the next August, they took in C with 400 dollars: 4 
the end of the year, they found there was a gain of 43) 
dollars. What share of the gain had each? . 

14. Gould and Davis entered into partnership for on 
year. Gould’s stock, at first, was only 500 dollars, bi 
at the end of 5 months he put in 150 dollars moré 
Davis’s stock, at first, was 600 dollars, but at the end o| 
9 months he took out 200 dollars: at the end of the yeail 
it was found they had gained 682 dollars 50 cents. Whe 
was the gain of each partner ? ‘fi 

15. Three farmers hired a pasture at 60 dollars 5) 
cents for the season. A put in 5 cows 4} months, J 
8 cows 5 months, and C 9 cows 64 mouths. Whe 
rent did each pay? ti 

16. A and B hired a coach in the city, to go 40 | 
for $20, with liberty to take in two more passengers) 
When they had ridden 15 miles they admitted C; and of 
their return, within 25 miles of the city, they admitted D. 
As each person is to pay in proportion to the distanes 
he rode, it is now required to settle the coach hire justly 
between them. 

17. Messrs. Howard, Bender, Dorr, and Tremeri, 
were partners for 2 years, under the firm of Joseph 
Howard and Co. When the firm commenced business’ 
Howard’s stock was 6000 dollars, Bender’s 3500 dollars’ 
Dorr’s 2800 dollars, and Tremere’s 1700 dollars. A’ 
the end of 8 months, Howard withdrew 3000 dollar: 
from the firm; after trading 10 months, Tremere added 
1300 dollars to his forraer stock; at the end of the firs’ 
year, Bender withdrew 800 dollars. At the close +f the 
two years; they had gained 3608 dollars 40 cents. How 
much was each partner’s share of the gain? # 


| 


| 
) 
! 


} 
eo 
| XXI. BANKRUPTCY. 121 
| 
ia , 
i; XXI. 
BANKRUPTCY. 


In the course of mercantile business, it happens not 
,unfrequently, that a merchant, cither from misfortune or 
_ imprudence, becomes insolvent, and what property he has, - 
yis distributed among his creditors in proportion to their 
{fespective dues. 

Questions in bankruptcy are performed on the same 
| principle with those in partnership: we first ascertain the 
amount of the bankrupt’s property, then the amount of 
his debts, next what per cent. he pays, and then multiply 
the sum due to each creditor by the decimal expressing 
the per cent. which the debtor pays. 


1. A man failing, owed the following sums: to A 120 
dollars 68 cents, to B 150 dollars 75 cents, to C 310 
collars 32 cents, to D 208 dollars 25 cents; and his whole 
‘property amounted to only 632 dollars, which was divided 
among them in proportion to their respective demands. 
‘How much did each receive? 

2. If the money and effects of a bankrupt amount to 

$361 dollars 74 cents, and he is indebted to A in the 
‘sum of 1782 dollars 24 cents, to B 1540 dollars 76 cents, 
and to C 2371 dollars 17 cents, how much will each 
of them receive ? 
3. A person failing in trade, owed A 539 dollars, B 
756 dollars 80 cents, C $54 dollars 16 cents, and D 
1200 dollars; his property amounted to 837 dollars 49 
cents, which was distributed among them in proportion 
to their several demands. How much did each creditor 
lose by the failure ? 

4, A bankrupt owes A 813 dollars 74 cents, B 3678 
(aeilars 46 cents, C 1840 dollars 40 cents, D 117 dollars 
80 cents, and E 814 dollars 60 cents, his whole proper- 
ty, worth 4029 dollars 30 cents, he gives up to his credi- 
tors. What per-cent. does he pay, and how much does 
each creditor receive ? 


| 


11 


122 ARITHMETIC. 


XXII. 


ASSESSMENT OF TAXES. 


TAXES are imposts paid by the people for the suppor 
of government. They are assessed on the citizens i 
proportion to their property; except the poll tax , which 
is assessed by the head without regard to property. | 

An accurate inventory of all the taxable property of 
every citizen is indispensable to a just assessment of 
taxes, and is the first thing to be obtained. 

When a tax is to be assessed on property and polls, we: 
first ascertain the amount which the polls pay, and deduet 
it from the sum to be raised; then apportion the remain: 
der according to each man’s property. " 

To effect the apportionment, we find what per cent. of 
the whole property to be taxed, the sum to be raised, 
is; then multiply each man’s inventory by that per cent. 
expressed in decimals, and the product is his tax. 


1. A tax of four hundred and fifty dollars is to be as- 
sessed on a parish, in which there are 40 rateable polls. 
Of this tax, that to be assessed on the polls amounts to 
50 dollars; leaving 400 dollars to be assessed on the real) 
and personal property of the parish, which by inventory, 
is 40000 dollars. What must a parishioner pay, whose 
real estate in the inventory is 700 dollars, persons) prop- 
erty 150 dollars, and who pays for one poll? | 

400 dollars is one per cent. (.01) of 40000 dollars. | 

Real estate’ 700 
Personal estate 150 


i 
| 
| 
| 


Total property 850 
Ol 
Tax on property 8.50 | 
Poll tax 1.25 1 
Total tax $9.75 75 Ans. ‘| | 
2. An incorporated town, in which the real and person 
property was valued in the inventory at 72856 dollars, and 


| 
; — 
‘ 


| &; 3 
Sex. ASSESSMENT OF TAXES. 123 


_ in which there were 154 rateable polls, voted to raise 
_ 1285 dollars 34 cents by taxation. ‘The tax on each poll 
was 1 dollar 25 cents. How much did A pay, whose 
_ real estate stood in the inventory at 2146 dollars, personal 
} Property at 224 dollars, and who paid for 1 poll ? How 
_much did B pay, whose real estate in the inventory was 
, 1000 dollars, his Byer property 140 dollars, and who 
| paid for 2 polls? How much did C pay, whose real 
,estate was valued at 785 dollars, persone property at 103 
dollars, and who paid for 3 polls 2 How much did D 
pay, whose real estate in the inventory was 4000 dollars, 
,personal property 478 dollars, and who paid for 1 poll ? 
/How much did E pay, who had no real estate, whose 
personal propery was valued at 250 dollars, and who 
_ paid for 4 polls? How much did a single woman pay, 
whose real estate was valued at 500 dollars, and her per- 
,sonal property at 120 dollars ? 
__ Assessors find it most expedient to make a table, which 
shall exhibit at once, the tax on all sums, from $1 up to 
any amount required. ‘The table is made by multiplying 
the per cent. which the tax amounts to, by the several 
numbers, 1, 2, 3, 4, and so on. 
_ The following is a table of taxes to be made when 14 
per cent. is to be raised on the valuation of property. 


$1 pays .015 $20 pay .30 $200 pay $3.00 
oe. .03 BO ie. ae 300...6¢ 4:50 
“6 045 40 “ .60 400 ‘* 6.00 
SRG eb a SO hz BOO. 68 2.50 
mee .075 60‘; .90 600 ‘* 9.00 
ef . .09 ZOi6 As O85. ahi ZOO) 6610.50 
ae 105 80 ‘ 1.20 800 ‘ 12.00 
ae OG al: 36 900 ‘¢ 13.50 
ee 135 100 ‘* 1.50 1000 ‘* 15.00 
ee SS. 15 | 


_ 3. By the above table, what would be the tax on 
$6425 real estate, and $2346 personal estate ? 

\@ 4. By the above table, what would be the tax of a 
freeholder, whose real estate is valued at $9842, and 
“personal estate, at $15066; poll tax &1.95? 


———— eS 
= —— ET 


124 ARITHMETIC. XXII, 


XXIII. 


GENERAL AVERAGE. 


Whenever a ship is in distress, and the master delib-| 
erately makes a sacrifice of any part of the lading, or of 
the ship’s furniture, masts, spars, rigging, &c. for the) 
preservation of the rest, all the property on board, which’ 
is saved by the sacrifice, must contribute towards the 
value of what is thus sacrificed. The contribution is! 
called a GENERAL AVERAGE ; and the property sacrificed 
called the jettison. 

Ifa vessel is accidentally stranded, and by extraordis’ 
nary labor and expense is set afloat, and enabled to com- 
plete her voyage’with the cargo on board, the expense! 
bestowed for this object also becomes a subject for general 
average. +i 

When a vessel has been forced by accidents, arising | 
from the perils of the sea, to enter a port in order to re- 
pair, all the charges incurred in consequence, together! 
with the wages and provisions of the master and crew{ 
during the delay, are brought into a general average. [ 

The contributory interests are the ship, the cargo, and’ 
the freight; and these must be cleared of all charges at-| 
tached to them, before the average is made. 

The contributory value of freight, in the ports general- | 
ly, is ascertained by deducting one third of the gross 
freight; in New York, however, one half is deducted. 
This deduction is made for seamen’s wages. 

In computing a general average for masts, rigging, &c. 
a deduction of one third is made from the cost of ree | 
placing them; because the new articles are supposed to | 
be so much better than the old ones. 

Particular average is nothing more than a partial loss 
and is borne wholly by the owner of the property damage 
tn making a general average, the partial loss, or partic 
average, is deducted from the original value of the dam 
aged property, and the remainder contributes to the 
general average. 


XXII. GENERAL AVERAGE. | 195 


The jettison contributes to a general average; other- 
wise its owner would not share im the general loss. 


1. The brig Ceres sailed from Gottenburs on the 
94th. of August 1833, bound for Boston with a cargo of 
iron and steel. She suffered considerable damage from 
tempestuous weather, and arrived in Boston harbor Oc- 


. tober 13th., where she got aground on Williams’s Island, 
_ and was obliged to discharge part of her cargo in lighters, 


. in order to get the vessel off. After this she was moored 
in safety at a wharf. 


The expense of lightening the vessel, to get her afloat, 


was 106 dollars and one cent, and was borne by a general 


average. 
From the surveyor’s report, it appears that the damage 


 gustained by the vessel on the voyage was 1195 dollars 


| James Fullerton & Co. $1.46 


73cents. The premium for insurance was 304 dollars 


27 cents. 
Contributory interests. 2 
- Vessel, valued at $ 8000. 
Less, damage and premium, 1500. 6500. 
Breight, gross amount, 1276.96 


, aes 5 4, as usual, for seamen’s wages, 425.65 851.31 


_ Cargo, shipped by 
James Fullerton & Co. $273.82 


Joshua Crane 626.67 

John Bradford 10378.48 

Wm. Parsons 1144.32 12423.29 
729691, = .00536-+ ‘i $19774.60 


Apportionment of contribution. 


Vessel, (6500 by .00536) pays 34.84 
Freight, ob Bie tare at tg 4.56 
Wargo, 12423.29 ‘* “* ° 66.61 $ 106.01 


shua Crane 37 
ohn Bradford 55.64 
m. Parsons 6.14 


$66.61 


1 hg 


126 ARITHMETIC. XXIIl” 


’  * 
2. ‘The schooner Crescent, on her passage from East- 


port to New York, sustained so much damage, that she 
was obliged to put into Plymouth to repair. The ex-" 
penses incurred by putting mto this harbor, viz. the pilot | 
age, protest, dockage, commission, wages and provision | 


of the master and crew while in harbor, amounting to 


$73.18, were paid by a general average, made in New | 


York, on the arrival of the vessel in that port. 


The vessel was valued at $4500, and the premium and_ 


repairs were estimated at $900. The gross freight was 
$153. } 
Cargo, shipped by E. Foster $ 600. 


Greason and H aughton 240. ‘ | 


Gold and Tucker 210. 
‘ Bucknam and Gunnison 400. 
Samuel Whaler 160 


Buck and Hammond 221.37 
What per cent. of the contributory interests was the 


general average? How much did each of the interests, _ 


and each of the shippers pay ? 


3. Ship Coral, on her passage from Boston to New 
Orleans, grounded at the bar of the river Mississippi, 
threw overboard part of her cargo to lighten, when near 
the breakers; broke an anchor, anchor stock, and wind- 
lass by strain in heaving off, and took a steamboat to 
bring her into port, while in this disabled situation. 

Statement of loss to be made up by a general aver- 
age. 

A. Howard’s goods, thrown overboard, 

Expense of steamboat, 

Freight lost in consequence of jettison, 

Damage to cable in heaving off, 

Anchor broke and lost, 

All other damage, 

Protest $14. Adjusting average $50. 


Agent’s commission 5 per cent. 
‘ Amount to be made up by the average, 


| 


i CUSTOM-HOUSE BUSINESS. 127 
Contributory interests. 
“Ship valued at N. Orleans, in cash, $ 11000. 
Freight, gross amount $862.48, less 4 4, 574.99 


yee shipped by Bridge & More $18135. 
Haw & Mears 17000. 
Gray & Bellows 14680. 
James Russell 3670. 
| A. Howard 925. 
| bt | #54410. 54410. 
Amount of contributory interests, $$ 65984.99 
What is the loss per cent.? How much does the ship, 


_ how much does the freight, and how much does each of 
the shippers contribute to the general average ? 


XXIV. 
CUSTOM-HOUSE BUSINESS. 


In every port from which merchandise is exported to 
foreign countries, and into which foreign merchandise is 
imported, there is an establishment under- the direction 
of the government, called a Custom House. The object 
of this establishment is, to execute the laws of the United 
States in the collection of duties imposed on certain arti- 
cles of imported merchandise, and on the tonnage of 
vessels employed in commerce. 
In order to secure the exact collection of duties, the: 
law provides, that the cargoes of vessels employed in 
foreign commerce, shall be inspected, and weighed or 
gauged by the custom-house officers. In the custom- 
house weight and gauge of goods, certain allowances are 
- made on account of the box, cask, bag, &c. containing 
jae goods, and on account of leakage, breakage, &c. 


ss 
a ALLOWANCES 
i Draft is an allowance made from the weight of each box, 


bag, cask, &c. of goods, on account of probable waste. 
i , 
7. 


; 


| 


| , 
128 ARITHMETIC. XXIV. 
| 


Tare is an allowance made for the weight of the box, 
bag, cask, &c. containing the goods. é| 
The whole weight of any parcel of goods, including 
the weight of the box, bag, cask, &c. containing the. 
goods, is called the gross weight. 


tare have been deducted, is called the neat weight. 
The allowance for draft is stated in the following table. 
Ona single box, &c. weighing 1 ewt. or 112 Ib. 1]b. 
** weighing above 1 cwt. and under. 2 cwt. 2 |b. 
“c  “*  hwweighing §=2 cwt.and under 3 cwt. © 3 Ih,) 


4 


Cee yi ee 3 cwt. and under 10 cwt. 4 Ih.| 
Celeste 66 10 cwt. and under 18 cwt. 7 Ib.° 
come ce «6 18 cwt. and upwards, 9 Ib. 


Observe, that the tare is computed on the remainder 
of any weight, after the draft has been allowed: and in 
casting tare, any remainder, which does not exceed half a. 
pound, is not reckoned; but, if it exceed half a pound, it. 
is reckoned a pound. 

The tare on sugar in'casks, (exceptloaf) is 12 per cent. 


| 
| 
| 
The weight of any parcel of goods after the draft and 
| 


a 


*¢ on sugar in boxes, - - 15 per cente} 
** on sugar in bags or mats, - - § per cenie| 
‘* on cheese in hampers or baskets, 10 per cent. ' 
** on cheese in boxes, - - 20 per cent 

‘¢ on candles in boxes, - - 8 per cent. | 
*€ on ehocolate in boxes, = - - 10 per centm, 
‘¢ on cotton in bales, - - - 2 per cent. 

*¢ on cotton in ceroons, - - 6 per cent. | 
** on glauber salts in casks, - - 8 per cent. | 
** on nails in casks, = - - - 3 per centm, 
‘¢’ on pepper in casks, - - - 12 per cent, , 
** on pepper in bales, - - - 5 per centm) 
** on pepper in bags, - = -- - 2 per centy | 
‘* on sugar candy in boxes, - - 10 per cent: | 
‘* on soap in boxes, - - - 10 per cent. | 
‘‘ on shot incasks, - «= - 3 percent: | 
‘* on twine in casks, - ~, -. 12 per cents 

‘* on twine in bales, — - . - 3 per ceng 


¥ 
j 
: 
' 


‘XXIV. CUSTOM-HOUSE BUSINESS. 199 


Tare on all other goods paying a specific duty, is 

allowed according to the statement of the same in the 
‘invoice of the goods, which is considered the actual 
weight of the box, bag, cask, &c. 
__ The importer may always have the invoice tare allowed, 
‘if he make his election at the time of making his entry, 
and obtain the consent of the collector and naval officer. 
| For leakage, 2 per cent. is allowed on the gauge, on 
all merchandise in casks paying duty by the gallon. 

For breakage, 10 per cent. is allowed on all beer, ale, 
and porter in bottles, and 5 per cent. on all other liquors 
in bottles; or the importer may have the duties computed 
on the actual quantity by tale, if he so chooses at the 
time of entry. 

_ The common-size bottles are estimated at the Custom- 
house to contain 22 gallons per dozen. 


1. What is the neat weight of 40 hogsheads of sugar 
weighing gross 8cwt. 3qr. each; draft and tare as in the 
tables ? . 

: ) ' 39200 Ib. gross weight. 
40 K 4== 160 lb. draft. 
39040 
12 per cent. of 39040 is 4684.8 4685 Ib. tare. 
34355 |b. neat weight. 


2. What is the neat weight of 25 bags of pepper, weigh- 
ing gross Iewt. each; draft and tare as mm the tables ? 

3. Find the neat weight of 6 chests of Souchong tea, 
weighing gross 98]b. each, tare 22lb. per chest. 

4. Find the neat weight of 12 casks of raisins, weigh- 
ing gross 130lb. each; draft as in the table, tare 12|b. 
per cask. 

5. What is the neat weight of 8 chests of green tea; 
famess weight 102]b. each, tare 20]b. per chest ? 

6. What is the neat weight of 9 bags of coffee, weigh- 
ing gross 114 lb. each, draft as in table, tare 2‘per cent. ? 

_ %. What is the neat weight of 4 casks of glauber salts, 
gross weight as follows; the first 1501b.; the 2d. 175|b.; 
3d. 228lb.; 4th. 264lb.; draft and tare as in tables ? 


130 “a ARITHMETIC. XXIV. 
; 


8. What is the neat weight of 4 hogsheads of madder 
weighing gross ll ewt. 2qr. each; draft being allowed a 
in the table, tare lewt. 2 qr. per cask ? . 


; 

: 

: : 
DUTIES, 1 

The duties paid on goods imported from foreign coun 
tries into the United States, are either ad valorem o. 
specific. | 
The ad valorem duty is a certain per cent. of the 
actual cost of the goods in the country from which they, 
are brought. | 
The specific duty is fixed at a certain sum per ton, 
hundred weight, pound, gallon, square yard, &c. 
Observe that the allowances for tare, draft, &c. are. 


to be made, before the duties are computed. 7 


9. What is the duty on an invoice of silk goods, which 
cost in Canton 4836 dollars, at 10 per cent. ad valorem? 

10. What is the duty on an invoice of woollen goods, | 
which cost in England 5729 dollars, at 44 per cent. 
ad valorem ? - 

11. Compute the duty on 6 boxes of chocolate, weigh-. 
ing gross Lcwt. per box; draft and tare as in the tables; 
duty 4 cents per Ib. ; 

12. Cast the duty on 12 boxes of Windsor soap; | 
gross weight 841b. per box; cost in England 1 dollar } 
per lb.; tare as in the table; duty 15 per cent. a 

13. Calculate the duty on 5 boxes brown Havana 
sugar; gross weight as follows; the first, 7ewt. 2Qqr.; 
2d. 8cwt. 3qr.; 3d. 9ewt. 1gr.; 4th. 10cwt. Sqr. 
20Ib.; 5th. Ll cwt. qr. 141b.; draft and tare as in the | 
tables; duty 23 cents per Ib. | 

14. What is the duty on a cargo of 148 tons of iron, 
at 30 dollars per ton? mi 4 

15. Compute the duty on 4 pipes of wine; allowance | 
for leakage as in the table; duty 73 cents per gallon. | 

16. Cast the duty on 10 gross of London porter; _ 
allowance for breakage as in the table; duty 20 cents 
per gallon. 


Se —_ 


ie 


x 
RXV. RATIO. a” 131 


17. What is the duty on 10 boxes of Spanish cigars, 

containing 1100 each; duty $2.50 per 1000? 

18. Compute the duty on 4 casks of Rochelle salts, 
invoiced at $10 per cwt.; gross weight of Ist cask 1 ewt. 

“2qr. 12\lb.; 2d. lewt. 1 qr. 17lb.; 3d. 2 cwt. 3qr. 7 Ib.; 

4th. 4cwt. Iqr.; draft as in table; tare 8 per cent.; 

duty 15 per cent. ad valorem. 


| | XXV. 
RATIO. 


~Rario is the mutual relation of two quantities of the 
_ game kind to one another. 
_ By finding how many times one number is contained 
in another, or what part one number is of another, we 
obtain their ratio. Thus, the ratio of 2 to 4 is 2, because 
'2is contained 2 times in 4; and the inverse ratio is 7, 
because 2 is 3 of 4. Boththese expressions of the ratio 
of 2 to 4 amount to the same thing, which is, that one of 
the numbers is twice as great as the other. 
By the ratio of two quantities is meant only their rela- 
tive magnitude; for, notwithstanding the absolute magni- 
tude of 2 pounds and 8 pounds is much greater than 
that of 2 ounces and 8 ounces, yet the rel tive magnitude 
or ratio of the two latter is just the same with that of the 
two former; because, 2 ounces are contained just as many 
times in 8 ounces, as 2 pounds are in 8 pounds; or, 2 
ounces are just as great a part of 8 ounces, as 2 pounds 
of 8 pounds. 
It is evident that only quantities of the same denomi- 
nation can have a ratio to one another; for it would be 
absurd to inquire how many times 1 dollar is contained in 
4rods, or what part of 4 rods 1 dollar is. 
| A ratio is denoted by two dots, similar to a colon: thus, 
3:9 expresses the ratio of 3 to 9. The former term 
of a ratio is called the antecedent, and the latter the con- 
sequent. Thus 6 : 12 expresses the ratio of 6 to 12, in 
which 6 is the antecedent, and 12 the consequent. 


2 | 
132 ARITHMETIC. XXV. 


Since a ratio indicates how many times one number 
is contained in another, or what part one number is of 
another, it is a quotient resulting from the division of one 
of the terms of the ratio by the other, and may be ex- 
pressed in the form of a fraction: thus, the ratio 6: 3 may 
be expressed by the fraction 2, or conversely $. } 

When any two numbers are multiplied, each by the 
same number, the ratio of the products is the same with 
the ratio of the multiplicands. Thus, take 3 : 6, and multi- 
ply the antecedent and consequent, each by 5, and the 
products 15 and 30 have the same ratio with 3 and 6; 
that is, 15 is contained just as many times in 30, as 3 is, 
in 6; or 15 is the same part of 30, that 3 is of 6. | 

Also, if two numbers be divided, each by the same 
number, the ratio of the quotients is the same with the 
ratio of the dividends. Thus, take the ratio of 9 : 18, and 
divide each term by 3, and the quotients 3 and 6 have 
the same ratio with 9 and 18; because 3 is contained as. 
many times in 6, as 9 is in 18; or 3 is the same part of. 
6, that 9 is of 18. 2 | 

A ratio resulting from the multiplication of two or more, 
ratios together, that is, the antecedents into the antece-. 
dents, and the consequents into the consequents, is called. 
a compound ratio. Thus, 6 : 48 is the compound ratio of | 
1: 2,3: 4, and 2: 6; because 6 is the product of all the) 
antecedents, and 48 of all the consequents. - This is om 
pressed in fractions with the word ‘‘ of”? between them: 
thus, making the antecedents the numerators, 4 of 3 of 25, 
making the consequents the numerators, ? of 4 of §. | 

‘T'wo ratios may be equal to one another, as well as two. 
quantities. The equality of two ratios is denoted by the. 
sign placed between them; thus, 2: 4—3: 6 signifies | 
that the ratio of 2 to 4 is equal to the ratio of 3to 6. 


PROPORTION. . 4 


The equality of 2 ratios is called a PRopoRTION, and — 
the terms are called proportionals; and in a proportion, | 


the first and fourth terms, that is, the antecedent of the first _ 


5 es 


| 
| 


XXV. PROPORTION. 193 


Yatio and the consequent of the second, are called the 
extreme terms; and the second and third terms, that is, 
the consequent of the first ratio and the antecedent of 
the second, are called the mean terms. Thus, in the pro- 
' portion 3: 9=4 : 12, 3 and 12 are the extreme terms, 
| 9 and 4 the mean terms. 

If the antecedent of the second ratio be the same with 


| the consequent of the first, the terms are in continued 
1 


“proportion. Thus, 3, 9, and 27, are ia continued propor- 
' tion, because 3: 9==9 : 27. 

_ Since the equality of two ratios constitutes a propor- 
tion, we can easily decide whether any four numbers be 
_in proportion, by bringing the fractions expressing the two 
-ftatios to a common denominator; for then, if the numbers 
be proportionals, the numerators also will be equal to one 
another. 

Take the numbers 4, 2, 6, 3; if we make the conse- 
quents the numerators, the fraction expressing the ratio 
of the two first in the series is 7, and that expressing the 
tatio of the two last is 2. These fractions, when reduced 
to acommon denominator, become 3% and 343 and this 
equality of the two fractions expressing the two ratios, 
proves that the four numbers are proportionals; for, if 
the four numbers were not in proportion, the fraction 
expressing the first ratio not being equal to the fraction 
@xpressing the second ratio, the numerator of the one 
would not be equal to the numerator of the other, when 
-feduced to a common denominator. 

_ Again, let us take the same numbers, 4, 2, 6, 3, and 
“make the antecedents the numerators of the fractions 
"expressing the ratios: thus, $ and §. These fractions when 
reduced toa common denominator, are 2 and 17, which, 
being equal, prove the four numbers to be proportionals. 
We see, therefore, whether we make the antecedents 
or consequents the numerators of the fractions expressing 
the ratios, that in both cases the equality of the ratios 
| proves a proportion among the four numbers; and in both 
_@ases the numerators are precisely the same; for in the 
first case the fractions are 43 and 42, and in the second, 
and ¥2, and these numerators, in both cases, are 
| 12 
oF 


* 
a 


q 


134 ARITHMETIC. xXxxyV 


products obtained by multiplying together the extrem) 
terms 4 and 3S, and the mean terms 2 and 6. 

These results prove, that, if four numbers be in pro. 
portion, the product of the two extreme terms is equal t 
the product of the two mean terms:.a principle of grea, 
practical utility, and the foundation of the ancient Rout 
or THREE. | 

It follows from what has been said, that the order of 
the terms of a proportion may be changed, provided they 
be so placed, that the product of the extremes shall be 
equal to the product of the means; because, wheneve: 
the product of the extreme terms of four numbers is equa! 
to the product of the mean terms, the numbers are pro: 
portionals: 

Take, for example, the proportion 3.:9==8: 9a 
3: 9==8: 24, and observe the dif- 3 .:8=9: 24 
ferent orders in which its terms may 24:8—9: 3 
be arranged. 24°; 8==8 ae 

That these changes do not disturb the proportion is 
evident; for the same numbers, which are the extreme 
terms in the first proportion, are the extreme terms in all 
the proportions; and the numbers, which are the mean 
terms in the first proportion, are the mean terms in all 
the proportions; therefore the products of the extremes 
and the products of the means must be the same in all 
the proportions. : | 


Again, the order of the above pro- 9: 324 BF 
portionals may be so changed, that 8: 3 =24:9) 
the mean terms shall become the ex- 9:24—3: 8 
treme terms, and the extreme terms 8:24-— 3 a 


the mean terms. | 

Since both the terms of a ratio may be multiplied or 
divided by the same number without altering the ratio, | 
it follows, that all the terms of a proportion may be mul-: 
tiplied or divided by the same number without disturbing | 
the proportion. Let us take, for example, the propor- 
tion 2: 46 : 12, and multiply each of the terms by 2, 
and we shall have the proportion 4: 8—=12: 24. If, 
instead of multiplying, we divide the terms of the same pro-| 
portion by 2, we shall have the proportion 1: 2=3: 6. 


| eV. PROPORTION. 135 


_ Either the two antecedents or the two consequents, in 
_ two equal ratios, may be multiplied or divided by the 


» same number without destroying the proportion; because 


_ the two ratios are increased or diminished alike, and 


#; therefore remain equal. Take the proportion 4: 16= 


ae 


» 6: 24, and multiply each of the antecedents by 2, and it 
_ will be 8 : 1612 : 24; if, instead of the antecedents, 

_ we multiply the consequents by 2, we have the propor- 

tion 4 : 326 : 48; if, instead of multiplying, we divide 

. each of the antecedents by 2, we have the proportion 
, 2: 16=3: 24; if, instead of the antecedents, we divide 
_ the consequents, we have the proportion 4 : 8==6 : 12. 

We may also multiply the antecedents and divide the 
conscquents at the same time, and vice versa, without 
_ destroying the proportion. If, for example, we take the 
_ proportion 3 : 6=9.: 18, and multiply each of the ante- 
cedents by 3, and divide each of the consequents by the 
| same number, we have the proportion 9 : 227 : 6; if 
| we multiply the antecedents by 3, and divide the conse- 
_ quents by 2, we have the proportion 9 : 327: 9; if 
| we divide the antecedents by 3, and multiply the conse- 
_ quents by 2, we have 1: 12=3: 36. 

Two or more proportions may be multiplied together, 
_term by term, and the products will be proportionals; 
for it is the same as multiplymg two equal fractions by 
two other equal fractions, the products of which will again 
_ be equal to each other. We give the following as an 
_ example. 3:46: 8 

2: 3=8 +12 
6 :12—48 : 96 

We may also divide one proportion by enother, term 
by term, with equal correctness of conclusion; for this is 
aly dividing two equal fractions by two other equal 
fractions, the quotients of which will again be equal. 
Take, for example, the proportion 24 : 3227 : 36, 
and divide it by the proportion 6 : 29: 3, term by 
_ term, and it gives 4 : 163: 12. 

A great variety of other changes may be made by 
differently multiplying, or dividing, or both; and such 
changes are frequently convenient in solving questions. 


136 ARITHMETIC. Xxv. 


The magnitudes of proportionals are changed witliout 
destroying the proportionality, when either the antece- 
dents, or consequents, or both, are respectively increas- 
ed or diminished by quantities having the same ratio; or 
when the two terms of either or of both ratios are re- 
spectively increased or diminished by quantities in the 
same ratio with themselves. We will make a few such 


changes in the proportionals 8, 6, 20, and 15. . 
8:6 =20 : 15 
84+6:6  =20+15 :15, or 14: 635 : 15! 
8—6:6  =20—15 :15, or 2:6= 5 : 15] 
8 : 8—6 ==20 : 20—15, or 8 :2=—90: 5 . 
8:8+6 =20: 20+15, or 8 :14==20 : 35] 
8+6 :8—6—=20+15 :20—15, or 14 :2=35:5 | 


Since the product of the extremes in every proportion | 
is equal to the product of the means, one product may | 
be taken for the other: now if we divide the preduct of | 
the extremes by one extreme, the quotient is the other | 
extreme; therefore, if we divide the product of the’ 
means by one extreme, the quotient is the other extreme: | 
for the same reason if we divide the’ product of the ex- | 
tremes by one of the means, the quotient is the other | 
inean; consequently, we can find any one term of a pro- | 
portion, when we know the other three. 2) 

To apply these principles to practice, let it be asked— 
If 64 yards of cloth cost 304 dollars, what will 36 yards 
cost? In the first place, the ratio of the two pieces of | 
cloth is 64 : 36; and secondly, the prices are in the same | 
ratio; that is, 304 dollars must have the same ratio to the | 
price of 36 yards, that 64 yards have to 36 yards. Now, 
if we put A. instead of the answer, we shall have the fol- 
lowing proportion, 64 : 86== 304: A, in which the pro- 
duct of the means is 10944, which, being divided by 64, 
one of the extremes, gives the quotient 171, the other | 
extreme, which was the term sought; therefore, 171 dok 
Jars is the price of 36 yards. 9 

Of the four numbers, which constitute a proportion, 
two are of one kind, and two of another. In the pre- 
ceding example, two of the terms are yards, and two | 
are dollars. | 


8 es 


| SRV. PROPORTION. 137 


If there are different denominations in the two first terms, 
they must both be reduced to the lowest denomination 
in either of them; and the third term must be reduced 
to the lowest denomination mentioned in it. Thus, if 
4 yards cost 18 shillings and 6 pence, what will 3 yards 
1 quarter 2 nails cost? Nails being the lowest denomi- 


nation in the two first terms, they must both be reduced 


_ to nails; pence being the lowest denommation in the 
third term, this term must be reduced to pence; and 


_ when thus reduced, the terms will make the following 


nas. nas. pence pence 


_ proportion; 64 : 54-222: A. The answer, when ob- 


tained, being in pence, must be reduced to shillings and 
pounds. In this question the answer is 15s. 774d. 

From the principles of ratio and proportion, which 
have been explained, we deduce the following rule for 
solving questions. 

RULE. Make the number, which is of the same kind 
with the answer, the third term; and, tf from the nature 
of the question, the third term must be greater than the 
fourth term or answer, make the greater of the two re- 
maining terms the first term, and the smaller the second; 
but, if the third term must be less than the fourth, make 
the less of the two remaining terms the first term, and the 
greater the second: then multiply the second and third 
terms together, and divide the product by the first term: 
the quottent will be the fourth term, or answer. 


1. If I buy 871 yards of cotton cloth for 78 dollars 39 
cents, what is the price of 29 yards of the same? 


871 : 29—78.39 : A The statements of this 
29 question may be read thus 

70551 —The ratio of 871 to 29 

15678 is equal to the ratio of 

_ Amer paid 78.39 to the answer. Or 

od 1)2273.31(2.61 Ans. thus—As 871 yd. is to 29 
1742 yd., so is $78.39 to the 

5313 answer. The operation 

5226 amounts to nothing more 

S71 than the multiplication of 


a 


138 ARITHMETIC. XXV.. 


2. If 14 yard of cotton cloth cost 42 cents, what will 
87 yards cost? . 
3. If T can buy 14 yard of cotton cloth for 61 pence, 
how many yards can I buy for £10 6s. 8d.? 
4. If I buy 54 barrels of flour for 297 dollars, what 
must I give for 73 barrels, at the same rate ? | 
5. If '7 workmen can do a piece of work in 12 days, 
how many can do the same work in 3 days? | 
6. If 20 horses eat 70 bushels of oats in 3 weeks,’ 
how many bushels will 6 horses eat in the same time ? | 
7. If a piece of cloth containing 76 yards cost 136 | 
dollars 80 cents, what is that per ell English? 
8. Ifa staff 4 feet long cast a shadow 7 feet in length, | 
on level ground, what is the height of a steeple, whose 
shadow at the same time measures 198 feet ? i 
9. How many yards of paper 23 feet wide, will hang 
a room, that is 20 yards in circuit, and 9 feet high? 
10. A certam work having been accomplished in 12° 
days by working 4 hours a day, in what time might it 
have been done by working 6 hours a day? 4 
11. If 12 gallons of wine are worth 30 dollars, what 


is the value of a cask of the same kind of wine, containing | 
313 gallons ? | 

12. If 83 yards of cloth cost 4 dollars 20 cents, what 
will 134 yards cost, at the same rate ? 

13. How many yards of cloth 3 yard wide, are equal | 
to 30 yards 14 yard wide? | 

14. If 7 pounds of sugar cost 75 cents, how many 
pounds can I buy for 6 dollars ? | . 

15. If 2 pounds of sugar cost 25 cents, and 8 pounds | 
of sugar are worth 5 pounds of coffee, what will 100 ‘ 
pounds of coffee cost ? 

16. A merchant owning ~ of a vessel, sold 3 of his | 
share for 957 dollars. What was the vessel worth at that 
vate ? 

17. A merchant failing in trade, owes 62936 dollars 
39 cents; but his property amounts to only 38793 dol- | 
lars 96 cents, which his creditors agreed to accept, and | 
discharge him. How much does the creditor receive, to | 

| 
| 
| 


whom he owes 2778 dollars 63 cents ? 


' 
, 


XXV PROPORTION. 139 


18. Bought 3 tons of oil, for 503 dollars 25 cents; 85 

gallons of which having leaked out. I wish to know at 
what price per gallon I must sell the residue, that I may 
neither gain nor lose by the bargain. 

19. If, when the price of wheat is 6s 3d. a bushel, 
_the penny loaf weighs 9 0z., what ought it to weigh, when 
wheat is at Ss. 25d. a bushel? | 
_ 20. If 15 yards of cloth ¢ yard wide cost 6 dollars 25 
cents, what will 40 yards being yard wide cost? 

21. What quantity of water must I add to a pipe of 
‘mountain wine, for which I gave 110 dollars, to reduce 
the first cost to 75 cents a gallon? 

22. Borrowed of a friend 250 dollars for 7 months; 
and then, to repay him for his kindness, I loaned him 300 

dollars. How long must he keep the 300 dollars, to 
balance the previous favor ? . 

23. If 44cwt. can be carried 36 miles for 5} dollars, 
-how many pounds can be carried 20 miles for the same 
money ? | 

24. A person owning 2 of a coal mine, sells 3 of his 

share for 570 dollars. What is the whole mine worth, at 
the same rate ? 

25. If the discount on $106, for a year, be $6, what is 

the discount on $477, for the same time? 
26. If, when the days are 132 hours long, a traveller 
perform his journey in 354 days, in how many days 
will he perform the same journey, when the days shal! 
be 11,7; hours long? 
27. A regiment of soldiers consisting of 976 men, is 
to be new clothed; each coat to contain 24 yards of 
cloth 13 yard wide, and to be lined with shalloon § yard 
wide. How many yards of shalloon will be required ? 
28. If 830 men can performa piece of work in 15 days, 
how many men would accomplish the same piece of 
work, in a fifth part of the time ? ‘ 
| 29. What is the value of 172 pigs of lead, each weighing 
| Sewt. 2qrs. 174]b., at 29 dollars 584 cents per fother 
of 194cwt.? 

_ 30. A merchant gave his note for 1831 dollars 75 
cents, payable in 8 months; but the holder of the note 


| 


L 
140 ARITHMETIC. XXV_ 


being pressed for money, the merchant paid it in | 
months. Allowing money to be worth 6 per cent. a year 
what sum was requisite to redeem the note ? | 
31. If A can mow an acre of grass in 52 hours, ant 
B can mow 14 acre in 93 hours, in what time can they 
jointly mow 8} acres ? 
32. How much cambrick may be bought for £8 6s. 
35,d., if 291 yards cost £186 2s. 41d.? 7 
33. Ifa staff 3 feet 3 inches, high casts a shadow 5} 
feet long, what is the height of the steeple of Park stray 
church, in Boston, which, at the same time, casts { 
shadow of 368 feet 6 inches in length? 
34. A and B hired a pasture for $49.50, in which, 4 
pastured 13 cows, and B 19. What must each pay? 
35. If 220 yards in length and 22 yards in breadth 
make an acre, what must be the breadth of a lot that i 
121 yards in length, to contain an acre? 
36. If 365 men consume 75 barrels of provision in ¢ 
months, what number of barrels will 500 men consume, 
during the same time? | 
37. If 19 yards of linen cost $14.25, what will 435.6 
yards come to, at the same rate ? | 
38. ‘The value of 8.25 pounds of pure silver being) 
$ 128, what is the value of 376.7848 pounds ? 
39. Suppose sound to move 1106.3 feet in 1 second; 
how many miles distant is a cloud, in which lightning is, 
observed 47.5 seconds before the thunder is heard ? i 
40. It has been found, that 100 cubic inches of alco- 
hol and 82.5 cubic inches of water, when mixed, mea- 
sures only 177.41 cubic inches. If, then, 125 gallons of 
_ alcohol and 103.125 gallons of water be mixed, how 
many gallons will the mixture compose ? i 
* | 


- COMPOUND PROPORTION. 


When proportion is applied to questions, in which the 
relation of the required quantity to the given quantity of | 
the same kind is traced through two or more proportions, — 
it is called coMPOUND PROPORTION. 


XXV. COMPOUND PROPORTION. 14) 

For example, 16 men dug a trench 54 yards long and 
6 yards wide, in 6 days. How many men of equal ability 
and industry will dig a trench of the same depth, 135 yards 
long, and 4 yards wide, in 8 days? ) 

In the above question, the number of men required 
depends upon three circumstances; viz. the length of the 
trench, its width, and the number of days in which it is 
to be dug. If we omit the consideration of all the cir- 
cumstances except the length, the question will be sim- 
\ply this—If 16 men dug a trench 54 yards long, how 
many men will dig one 135 yards long ?—which will make 
the following proportion; 54 : 12516: A, and the 
fourth term will be found to be 40 men. 

Secondly, we will consider the width; and, since the 
second trench is to be narrower than the first, the num- 

ber of men required will be proportionally less, and our 
second proportion will be the following; 5: 4—40: A, 
and the fourth term will be 32 men. 

Lastly, we will notice the number of days in the ques- 
tion; and, since the longer the ume allowed, the less will 
be the number of men required to do the work, we shall 
have the following proportion; 8 : 6382: A, and this 
gives 24 men for the fourth term, which is the answer to 
the question. 

We see in this solution, that 16 is multiplied by 135, 
and the product divided by 54; the quotient, being made 
the third term in the second proportion, is multiplied by 
4, and the product divided by 5; this last quotient, 
being, made the third term in the third proportion, ts 
nultiplied by 6, and the product divided by 8. The re- 
sult, therefore, would be the same, if 135 and 4 and 6 
were multiplied together, and their product multiplied by 
16, and this last product divided by the product of 54 
ind 5 and 8. The proportion may be thus arranged. 


54: 135 
| Bt eee io. A 
| 3) ie 6 
2160: 3240 =—16:A 


| _ If, instead of calculating the fourth term m each pro- 
»ortion, we only indicate the operation by a fraction, 


| 
142 ARITHMETIC. XXV 


we shall have, in the first of the foregoing proportions 


“*—— for the fourth term: taking this for the third tern 


5: 4==———: A, and the fourth term will be —7am 
taking this for the third term in the third proportion, wr 
16x 1354 


shall have the following, 8 : 6=—3,,5— : A, and the fourtl 


"11 1, 16x 135x4x6 ahd clad 
term will be ee, which is equal to 24, the number 


of men required. In this fractional expression, we see 
at once, that the product of all the second terms is muk 
tiplied by the third term, and that this product is divided 
by the product of all the first terms, and the quotient is 
the fourth term, or answer to the question. 

Hence we see, that questions in compound proportion 
willbe accurately solved by the following rule. | 

RULE. Make the number, which is of the same kind 
with the answer, the third term; of the remaining num- 
bers, take any two of a kind, and write one for a first 
term and the other for a second term, as directed in sim- 
ple proportion, then any other two of a kind, and so on, 
itll all are written. Lastly, multiply all the second terms 
together, and their product by the third term, and divide 
the result by the product of the first terms; the quotient 
will be the fourth term, or answer. 


41. A wall to be built to the height of 27 feet, was 
raised to the height of 9 feet by 12 men in 6 days; how 
many men must be employed to finish it in 4 days? 

27 


fate 


ee 


12 men is the third term, because 
3 the answer is to be in men. ie | 
9: 18 —1]9: A | Im stating the different lengths — 


43° 0 of the wall, the shorter is made the’ 
36: 108 —12: A |first term, because, the longer the i" 
12 wall, the greater will be the number | 
TeeGate of men required to build it. , 
36) 1296 (36 Ans. In stating the days, the less num _ 
108 . ber is made the first term, because, | 
216 the less the time, the greater must 


216 + be the number of men. j 


m ¢ 
f 


‘ 


XXV. COMPOUND PROPORTION. 143 


42. If 120 bushels of corn will serve 14 horses 56 
‘days, how many days will 94 bushels serve 6 horses ? 

_ 43. If a footman travel 130 miles in 3 days, when the 
days are 12 hours long, in how many days of ten hours 
jm length, can he travel 360 miles ? 

44. If 6 laborers dig a ditch 34 yards long in 10 days, 
ov many yards will 20 laborers dig in 15 days? 

45. If a garrison of 600 men have provisions for 5 
weeks, allowing each man 12 ounces a day, how many 
men may be maintained 10 weeks with the same provi- 
sions, if each man is limited to 8 ounces a day ? 

| 46. If 3 bushels and 3 pecks of wheat will last a 
family of 9 persons 22 days, in how many days will 6 
persons consume 5 bushels ? 

47. If 450 tiles, each 12 inches square, will pave my 
cellar, how many tiles must I have, if they are only 9 
inches long and 8 inches broad ° 
, 48. If12 ounces of wool make 24 yards of cloth 6 
quarters wide, how much wool is required for 150 yards 
4 quarters wide ? 

49. Ifa bar of iron 4 feet long, 3 inches broad, and 
15 inch thick, weighs 36 pounds, what will a bar weigh, 
that is 6 ft. long, 4in. broad, and 2in. thick ? 

50. If 6 men built a wall 20 feet long, 6 feet high, 
and 4 feet thick, nm 16 days, in how many days will 24 
i build a wall 200 feet long, 8 feet high, and 6 feet 
t Be. P 


Se 


. If 14 men can reap 84 acres in 6 days, how many 
i. must be employed to reap 44 acres in 4 days? 
_ 52. Aship’s crew of 300 men were so supplied with 
provisions for 12 months, that each man was allowed 30 
ounces a day; but after having been 6 months on their 
voyage, they find it will take 9 months more to finish it, 
and 50 of their number have been lost. It is required, to 
find the daily allowance of each man during the last 9 
months. 

53. A wall was to be built 700 yards long in 29 days; 
after 12 men had been employed on it for 11 days, it was 
found they had built only 220 yards. How many more 
ten must be put on, to finish it in the given time? 


| 


444 ARITHMETIC. XXV 


ad 


54. Ifthe transportation of 12cwt. 2qr. 6lb., 275 mile) 
cost $27.78, how far, at that rate, may 3 tons Ocwt) 
3qr. be carried, for $234.78? . 

55. A cistern 173 feet in length, 104 feet in breadt 
and 13 feet deep, holds 546 barrels of water. Then hoy 
many barrels will fill a cistern, that is 16 feet long, 7 fee 
broad, and 15 feet deep? | 


56. If 25 pears can be bought for 10 lemons, and 2 
lemons for 18 pomegranates, and 1 pomegranate for 4¢| 
almonds, and 50 almonds for 70 chestnuts, and 10° 
chestnuts for 24 cents, how many pears can I buy fo} 
$1.35? 
‘ 57. In how many days, working 9 hours a day, will 
24 men dig a trench 420 yards long, 5 yards wide, ant 
3 yards deep; if 248 men, working 11 hours a day, in 
days dug a trench 230 yards long, 3 yards wide, and {| 
yards deep ? | 

58. If the intere.t on 347 dollars for 34 years be 7%! 
dollars 87 cents, what will be the interest, at the same 
rate, on 527 dollars for 24 years ? 

59. What must be paid for the carriage of 4cwt., 39) 
miles, if the carriage of 8cwt.,.128 miles, cost 12 dollarg| 
80 cents? | 

60. By working 9 hours a day, 5 men hoed 18 acres| 
of corn in 4 days. How many acres will 9 men hoe, ai 
that rate, n 3 days, working 10 hours a day? | 

61. One pound of thread makes 2 yards of linen cloth,| 
5 quarters wide. ‘Then how many pounds of thread will! 
be required to make 50 yards of linen 2 yd. wide? | 

62. If 6 men, working 7 hours a day, mowed 28| 
acres of grass in 4 days, how many men, at that rate, 
will mow 16 acres in 8 days, working 6 hours a day? 7 | 

63. If 5 men can make 300 pair of boots in 40 days, 

‘how many men must be employed to make 900 pair m 
60 days? 

64. If 3 compositors set 154 pages in 2% days, how! 
many will be required to set 693 pages in 61 days? 

65. If 36 yards of cloth, 7 quarters wide be worth| 
$ 98, what is the value of 120yd. of cloth of equal tex- 
ture, but only 5 quarters wide ? | 


XXXVI. CONJOINED PROPORTION _— 145 


4 XXVI. 
CONJOINED PROPORTION. 


CoNJOINED PROPORTION—called by merchants, The 
Chain Rule—consists of a series of terms bearing a certain 
proportion to each other, and so connected, that a com- 


/ parison is instituted between two of the terms, through 
| the medium of all the others. 


The principles of this rule are fholuded 3 in proportion. 


The rule is chiefly employed in the higher operations of 
| exchange, arbitrations of bullion, specie and merchandise. 
_ For the purpose of elucidation, however, we propose the 
| following familiar example. 


If 3lb. of tea be worth 4lb. of coffee, and 6lb. of 


coffee worth 20]b. of sugar, how many pounds of sugar 
‘may be had for 9lb. of tea? 


This question, we know, may be solved by a statement 
m compound proportion; but the following is the solu- 
tion by conjoined proportion. 

Distinguish the several terms into antecedents and con- 
sequents, and connect them by the sign of equality in the 


| way of equations, as follows. 


First, enter on the right the given sum or term on 
which the operation is to be performed, (which in the 


foregoing question is 9lb. of tea) and call this the term 
_ of demand.. 


Secondly, on the left of this term, and a line lower, 


‘enter the first antecedent, which must be of the same 


kind or name with the term of demand, and equal in value 
to the annexed consequent. 

Thirdly, in the same manner, let the second antecedent 
be of the same name with the second consequent, and 
equal in value to the third consequent: and so on, for any 
number of terms. 

Fourthly, the terms being thus arranged, divide the 
product of the consequents by the product of the antece- 


dents, and the quotient will be the answer in the denomi- 


nation of the last consequent, or odd term. 
12 


146 ARITHMETIC XXXVI. 


9lb. of tea, term of demand. 


3lb. tea = 4b. of coffee. .| 
6lb. coffee —20]b. sugar, the odd term. | 
Hence AROS = Ge =40]b. sugar, the answer. | 


By the above example it will be seen, that in the ar- 
rangement of antecedents and consequents, each sort is 
entered twice, except that in which the answer is required, 
and which is called the odd term. — 4 

It should also be observed, that no two entries of the 
same denomination are in the same column; and, as they | 
are placed in the way of equations, it is evident that the 
quantities on each side, which are equal in value to one 
another, are cancelled in the operation; and, therefore, | 
the quotient or answer will obviously be in the denomi- 
nation of the last consequent, which is the odd term. 

This rule may be proved by reversing the operation; 
taking the answer as the term of demand, and making the 
first antecedent the last consequent or odd term, as follows. | 

40lb. of sugar. | 
20lb. sugar — 6]b. coffee. 
4 lb. coffee— 3]b. tea. 
Then, = 20. 1 O)h. of tea, the f. 
4x20 80 ) prcet. 
The operation may be abridged by omitting such num- | 
bers as are the same in both columns, whenever such 
instances exist. | 

When fractions occur, the most convenient method is 
to convert them into whole numbers. Thus, an antece- 
dent of '5 and a consequent of 9 may be changed (by 
multiplying both by 12) into 7 and 108, and the ratio will | 
not be altered. So 5 and 113 have the same ratio with | 
20 and 47. | 

The rule may be exemplified by a question in reduc- 
tion; thus,-— It is required to reduce 2 tons to ounces. 

2tons, term ofdemand. | 


Lton — 20ewt. | | 
lowt. == 4qr. | 
Igr.. =) 28]b. | | 
1lb. == 16o0z. the odd term. 


16 X28 K 4x 202 
Then ~).1,1<1~ ==71680 ounces, the answer 


; 
/XXVI. CONJOINED PROPORTION. 147 


1. If 17]b. of raisins are worth 20 Ib. of almonds, and 
5Ib. of almonds worth 84]b. of figs, and 374 ]b. of figs 
worth 301b. of tamarinds, how many pounds of tamarinds 
are equal in value tc 42} Tb. of raisins ? 

__ 2. Suppose 100l]b. of Venice weight is equal to 70 1b. 

_of Lyons, and 60 Ib. of Lyons to 501b. of Rouen, and 

20ib. of Rouen to 25lb. of Toulouse, and 50lb. of 

Toulouse to 37 1b. of Geneva; then how many pounds of - 

Geneva are equal to 25lb. of Venice ? 

3. If 1 French crown is equal in value to S0 pence 

of Holland, and 83 pence of Holland to 48 pence Eng- 

lish, and 40 pence English to 70 pence of Hamburgh, 
and 64 pence of Hamburgh to 1 florin of Frankfort, 
how many florins of Frankfort are equal to 166 French 

oo. P 

If A can do as much work in 3 days as B can do 

in i days, and B as much in 9 days as C in 12 days, 

@nd C as much in 10 days as D in 8 days, how many — 

days’ work of D are equal to 5 days’ work of A? 

5. If 70 braces at Venice are equal to 75 braces at 
Leghora, and 7 braces at Leghorn are equal to 4 yards 
in the United States, how many braces at Venice are 
equal to 64 yards in the United States? 

6. A merchant in St. Petersburg owes 1000 ducats 
in Berlin, which he wishes to pay in rubles by the way 
of Holland; and he has for the data of his oper ation, the 
following information, viz. That 1 ruble gives 47} 

stivers; that 20 stivers make 1 florin; 24 florins 1 rix 
dollar of Holland; that 100 rix dollars of ‘Holland fetch 

142 rix dollars an Prussia; and that 1 ducat in Berlin is 

worth 3 rix dollars Prussian. How many rubles will 
pay the debt? 

7. If 94 piasters at Leghorn are equal to 100 ducats 
at Venice, and 1 ducat is equal to. 320 maravedis at 
Cadis, and 272 maravedis are equal to 630 reas at Lis- 
/ bon, and 400 reas are equal to 50d. at Amsterdam, and 
56d. are equal to 3 francs at Paris; and 9 francs are 
equal to 94d. at London, and 54d. are equal to 1 dollar 
nthe United States, how many dollars are equal to 800 


| 
| plasters ? 
| 
| 


148 ARITHMETIC. XXVIL 


XXVII. 


DUODECIMALS. 


DuopEciIMALs are compound numbers, the value of 


whose denominations diminish in a uniform ratio of 12. 
They are applied to square and cubic measure. : 
The denominations of duodecimals are the foot, (f.), 


the prime or inch, (‘), the second, (”), the third, er) 


the fourth, (’’’), the fifth, (’’”), and so on. Accordingly, 
the expression 3 1’ 7” 9” 6’” denotes 3 feet 1 prime 
7 seconds 9 thirds 6 fourths. % 
The accents, used to distinguish the denominations 
below feet, are called indices. q 
The foot being viewed as the unit, duodecimals pre- 
sent the following relations. 


i’ = of 1 foot. 

==, of 7, ef kfoot.: 2) se = qixz of 1 foot. 
1” = 75 of 7 of y5 of 1 foot... . == +,, of 1 foots 
1 = qq Of yg Of zg Of 35 of Lfoot. = ,1,-0f 1 foot. 


&c. 


Addition and subtraction of duodecimals are performed 
as addition and subtraction of other compound numbers; 
12 of a lower denomination making 1 of a higher. Mul- 
tiplication, however, when both the factors are duodeci- 
mals, is peculiar, and will now be considered. 


When feet are multiplied by feet, the product is in 
feet. For instance, if required to ascertain the superficial 
feet ina board 6 feet long and 2 feet wide, we multiply 
the length by the breadth, and thus find its superficial, or 
square feet to be 12. But when feet are multiplied by 
any number of inches [primes], the effect is the same as 
that of multiplying by so many twelfths of a foot, and 
therefore the product is in twelfths of a foot, or inches: 
thus a board 6 feet long and 6 inches wide contains 36 
inches, because the length being multiplied by the breadth, 
that is, 6 feet by 3 of a foot, the product is +5 of a foot, 


L 


nr 


| 
XXVII DUODECIMALS. 149 


or 36’=3 feet. When feet are multiplied by seconds, 
the product is in seconds: thus 6 feet multiplied by 6 
seconds, that is, ¢ of a foot by 3% of +5 of a foot, the 
product is 33°; of a foot, or 36” =3 inches. 
_ When inches are multiplied by inches, the product is 
'm seconds. ‘Thus, 6 inches multiplied by 8 inches, that 
18, 75 of a foot by +, of a foot, the product is 748; of a 
foot, or 48” —4 inches. When inches are multiplied by 
‘seconds, the product is in thirds. Thus, 6 inches multi- 
ed by 8 seconds, that is, =6; of a foot by 48 of 44 of a 
oot, the product is +43, of a foot, or 48°’ —=4 seconds. 
When seconds are multiplied by seconds, the product is 
in fourths. Thus, 6” multiplied by 8”, that is, 4% of +5 
of a foot, by + of 35 of a foot, the product is 548, 
of a foot, or.48’” = 4 thirds. 

This method of showmg the denomination of the pro- 
duct resulting from the multiplication of duodecimals by 
duodecimals may be extended to any number of places 
whatever; but sufficient has been said, to show that the 
product is always of that denomination denoted by the 
sum of the indices of the two factors.. 

Feet multiplied by feet, produce feet. 
Feet multiplied by primes, produce primes. 
Feet multiplied by seconds, produce seconds. 
Feet multiplied by thirds, produce thirds. 
WC. 
Primes multiplied by primes, produce seconds. 
Primes multiplied by seconds, produce thirds. 
Primes multiplied by thirds, produce fourths. 
&c 


Seconds multiplied by seconds, produce fourths. 
Seconds multiplied by thirds, poduce fifths. 
Seconds multiplied by fourths, produce sixths. 
&c. 
Thirds multiplied by thirds, produce sixths. 
Thirds multiplied by fourths, produce sevenths. 
&c. 
__ If we would find the square feet in a floor 6f. 4’ 8”sin 
Tength, and 4f. 6’ 5” in breadth, we should proceed as 
| follows. . 
13* 


) 


150 | ARITHMETIC. XXVIL | 


6f. 4° 8” | We began on the right hand, 
en and multiplied the whole multi- 


3 3, oy “a 4 the multiplier, then by the inches, 
05 6° 3” and lastly by the feet, and added 
“the results together, and thugi 
28f,1V 7” IY" 4” | obtained the answer. 


That the above answer is the true one, will appeak | 


very clearly from the following considerations. The 8 


equal to 3” and 4””; writing down the 4””, we reserve 
the 3’” to be added to the product of 4’ by 5". 4’ being 
is of a foot, and 5” being +3; of a foot, their product is 
773, of a foot, or 20”, to which adding the 3”, tnat were 
reserved, we had 23”, equal to 1” and 11”; we wrote 
down the 11”, and reserved the 1” to be added to the 
product of 6 feet by 5”. 6 feet being $ of a foot, and 5” 
being z#_ of a foot, their product is 4; of a foot, or 
30", to which we added the 1” reserved, and thus had 
31", equal to 2’ and 7”, both of which we wrote down. — 


Having completed the multiplication by the seconds, | 
we next multiplied by the inches: 8” being 7$; of a foot, — 
and 6’ being ;’5 of a foot, their product is 748, of a foot, — 
or 48==4"; we therefore put down a cipher in the place | 
of thirds, and reserved the 4” to be added to the product, | 


of inches by inches. 4 inches being +‘; of a foot, and 6 
inches +5 of a foot, their product is 74 of a foot, or 24”, 
to which we added the 4” reserved, making 28’ 2’ and 
4"; writing down the 4”, we reserved the 2’ to be added 
to the product of feet by inches. 6 feet being § of a foot, 
and 6 inches +5; of a foot, their product is 28 of a foot, 
or 36’, to which we added the 2’ reserved, making 38° 
= 3 feet and 2 inches, both of which we wrote down. 
Lastly, we multiplied by the feet in the multiplier. 8”, 
or 74g Of a foot being multiplied by 4 feet, or 4 of a foot, 
their product is 3%; of a foot, or 32”—=2’ and 8"; setting 
cown the 8”, we reserved the 2’ to be added to the pro- 
duct of inches by feet. 4’, or 74 of a foot being multiplied 


\ 


plicand, first by the seconds in | 


t 


| 


, 
| 
| 


| 
XXVIII. DUODECIMALS. 151 


by 4 feet, or ¢ of a foot, their product is 18 of a foot, or 
16’, to which we added the 2’ reserved, making 18’=—! 
foot and 6 inches; writing down the 6’, we reserved the 
1 foot to be added to the product of feet by feet. 6 feet 
being multiplied by 4 feet, their product is 24 feet, to 
which we added the 1 foot reserved, making 25 feet. By 
adding these three partial products together, we obtained 
the answer to the question. | 

Therefore, to multiply one number consisting of feet, 
inches, seconds, &c. by another of the same kind, we 
give the following rule. ’ 

RULE. Place the several terms of the multiplier under 
the corresponding ones of the multiplicand. Beginning 
on the right hand, multiply the several terms of the mul- 
tiplicand by the several terms of the multiplier successive- 
ly, placing the right hand term of each of the partial 
products under its multiplier; then add the partial pro- 
ducts together, observing to carry one for every twelve, 
both in multiplying and adding. The sum of the partial 
products will be the answer. 


Questions in duodecimals are very commonly performed 
by commencing the multiplication with the highest denom- 
ination of the multiplier, and placing the partial products 
as in the first of the two following operations. ‘The re- 
sult is the same, whichever method is adopted. The 
second operation, however, is according to the rule we 
have given, and is more conformable to the multiplica- 
tion of numbers accompanied by decimals. 


han hd S40 BF 
2f. 6’ 4° Ges a 
6. 5’ 9" 4! Qo” “10 i 4 WH 
1 is Sad 6” 1 mm! a 6” 

) 1’ Qo” 10” 4 iu 6 5! Qt" 

eae 1.6", 47 4°" Sf. 1’ 6% 4migm 


_ When there are not feet in both the factors, there may 
not be any feet in the product; but, after what has been 
said, there will be no difficulty in determining the places 
lof the product. 


* 


152 ARITHMETIC. ae 


4, Milvnly 46783" by af 6’. 

5. How many square feet are there in a house lot 4st. 
3’ in length, and 25f. 6’ in breadth ? 4 

6. What is the product of 10f. 4’ 5” by 7f. 8’ 6”? 

7. Calculate the square feet in an alley 44f. 2’ 9” long, 
and 2f. 10'3" 2" 4" wide. 

8. How many square feet are there in a garden 30f 
10’ 7" long, and 18f. 8’ 4” wide? o) 

9. What is the product of 24f. 10’8” 7” 5” by of 
4! 6 if ? 

10. Compute the solid feet in a wall 53f. 6’ long, 12 4 
3’ high, and 2f. thick. | 

11. The length of a room is 20 feet, its breadth 4 
feet 6’, and its height 10f. 4’. How many yards of paint=' 
mg are ‘there in its walls, deducting a fire place of 4f. by 
4f. 4’; and two windows, each 6f. by 3f. 2’? 

12. How many solid feet in a pile of wood 22f. 6! 
Jong, 12f. 8" wide, and 5f.’8’ high? 

13. How many yards of plastering i in the top and foil 
walls of a hall 58f. 8’ Jong, 21f. 4’ wide, and 13f. 9/ 
high; deducting for 2 doors each 7f. 6! high and 4f. | 
wide; for 7 windows each 6f. 2’ high, and Bf. 10’ widen 
for 2 fire places, each 3f. 6’ high, and 4f, ues and for 
a mop board 9 inches wide around the hall ? 

14. How many yards of papering in a room 17f. 8! : 
long, 16f. 9’ wide, and 12f. 6’ high; deducting for 2° 
doors each 6f. 6’ high, and 4f. wide; for a fire place - 
4f. 6' high and 3f. 10’ wide; for 3 windows each 5f. 6! | 
high and Bf. 8’ wide, and for a mop board 8 inches wide — 
around the room? 5 

15. How many yards of carpeting, yard wide, will be . 
required for a room 21f. 6’ long, and 18 ft. wide? | 

16. What will the plastering of a ceiling come to, at 10 
cents a square yard, supposing the length 21 feet 8 | 


( 


inches, and the breadth 14 feet 10 inches? | 
| 


SX VIL. INVOLUTION. 153 


XXVIII. 
INVOLUTION. 


InvoLuTIon is the multiplication of anumber by itself. 
The number, which is thus multiplied by itself is called 
he root. The product, which we obtain by multiplying 
number by itself, is called a power of that number. 
[he number involved is itself the first power, and is the 
oot of all the other powers. 

A number, multiplied once by itself, is said to be in- 
olved or raised to the second degree, or second power; 
auitiplied again, to the third degree, or third power; 
nd soon. For «xample, 3 X98 is raised to the second 
ower of 3, which is 9; 3X38 is raised to the third 
ower of 3, which is 27; &c. 

We distinguish the powers from one another by the 
umber of times, that the root is used as factor in the 
qitiplication of itself. Thus, 3X3 produces the sec- 
nd power of 3, because 3 is used twice as factor; 

x 3X3 produces the third power of 3, because 3 is 
sed three times as factor; 3X33 3 produces the 
surth power of 3, because 3 is used four times as factor; 
ad so on. 

A fraction is involved in the same manner by multiply- - 
ig it continually into itself; thus, the second power of 
sis 2X 3—7,; the third power is 7% xX 3=24; the 
yurth power is 24 X $==34; and so on. So also in deci. 
ials the second power of .2, is .2 X.2==.04; the third 
ower is .04 X.2=.008; the fourth power is .008 <.2 
=.0016; and so on. ? 

To: involve a mixed number, reduce it first to an im- 
roper fraction, or the vulgar fraction to a decimal, and 
ien involve it. Thus, 14 when reduced to an improper 
action, is 3, the second power of which {—2j; the 
urd power is 2733; &c. If, instead of reducing 13 
) an improper fraction, we reduce the vulgar fraction to 
decimal, we have 1.5, the second power of which is 
.25; the third power, 3.375; &c- 


| / 


| 


‘ 
r/ 


| 


The second power is commonly called the square; the: 
third power, the cube; the fourth power, the biquadrate, 
The other powers now generally receive no other thar 
numeral distinctions; as the fifth power, the sixth power, 
the seventh power, &c. In some books, however, the 
fifth power is called the first sursolid; the sixth power, 
the square cubed, or the cube squared; the seventh pow- 
er, the second sursolid; the eighth ~ower, the biquadnal 
squared; the ninth power, the cube cubed. | 

The powers of 1 remain always the same; because, 
whatever number of times we multiply 1 by itsclf, the 
product is always 1. e 

A power is sometimes denoted by a number, placed at 
the right hand of the upper part of the root; thus, 52 db 
notes the second power ef 5, which is 25; 42 denotes the! 
third power of 4, which is 64; 94 denotes the fourth 
power of 9, which 6561; &c. The number, thus used 
to denote the power, is sometimes called the exponent 
and sometimes the index. But the use of these exponents 
or indices in arithmetic is very limited; they belong chiefly 
to algebra. 7 

We will now make afew observations on the result! 
arising from the multiplication or division of one power 
by another. To illustrate this subject, we will take UP 
number 3; we must here observe, however, that since 
every number is the first power of itself, the exponent 1 
is never expressed; so that 3 and 3! mean the same thing; | 
the exponent | being always understood, when no expe 
nent is expressed. Now 3 multiplied by 3 produces 
the second power of 3, which may be thus expressed, | 
3* X 31 == 37; so also 32 X 3133; and 33x 3} == Ga 
&c. We have here expressed the exponent 1 for the | 
purpose of showing that we obtain the exponent of the. 
product or power produced, by adding together the 
exponents of the factors or powers used in producing it. | 

1 
| 


154 ARITHMETIC. XXVITL 


‘ 


Hence the second power of any number multiplied by the 
second power of the same number produces the fourth 
power of that number; thus, 32 x 32==3?: the third pow-- 
er multiplied by the third power gives the sixth power; | 
as 2° X 2°==2°: the fourth power multiplied by the see- | 


| 
| 
i 


XX VIUIT. INVOLUTION. 155 


ond power gives the sixth power; as 2+ x 2°==2°: the 

fourth power multiplied by the fourth power produces the 

eighth power; as 34 x 34= 838: the third power multiplied 

by the third power, and the product again by the third 

power gives the ninth power; as 23 x 23 « 33—=9?, 

_ Division being the reverse of multiplication, it is evi- 

* that if we subtract the exponent of the divisor from 
e exponent of the dividend, the remainder is the expo- 

aent of the quotient. For example, if we divide the fifth 

power by the third power, the quotient is the second 

power; as 3°+3°== 3°: if we divide the ninth power by 

the sixth power, the quotient is the third power; as 

3? 6° 6°: if we divide the ninth power' by the eighth 

power, the quotient is the first power; as 69+ 6°9=6. 

. What is the third power of 12? 

. Find the fourth power of 11. 

Raise 13 to the fifth power. 

What is the square of 27? 

. What is the square of .27? 

Raise .7 to the fourth power. 

. What is the eighth power of .2? 

. What is the third power of .1? 

. What is the square of 2 ? 

10. What is the.cube of 3? 

11. Involve 4; to the third power. 

12. Raise 2 to the fourth power. 

_ 18. What is the square of 30} ? 

_ 14. What is the biquadrate of 31 ? 

15. Involve 1.1 to the fifth power. 

16. Raise 20% to the fourth power. 

17. Raise 8.2 to the third power. 

18. What is the fourth power of 17? 

19. Divide 7° by 7%, and write the quotient. 

_ 20. Multiply 8° by 8, and write the product. 

21. What is the quotient resulting from 57+ 54? 

22. What is the product resulting from 63 X6? 

23. Multiply 9? by 9, and write the product. 

24. Divide 41° by 4°, and write the quotient. 

25. What quotient results from 19°+197? 

| 

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| XXVIII EVOLUTION. 157 


EVOLUTION. 


Evo.Lurion is the reverse of involution; for in involu- 
_ tion we have the root given, to find the power; but in 
_ evolution we have the power given, to find the root. 

Power and root are correlative terms; for, as 4 is the 
square of 2, 2 is the square root of 4; as 8 is the cube 
of 2, 2 is the cube root of 8; as 16 is the biquadrate of 
_ 2, 2 is the biquadrate root of 16; as 32 is the fifth pow- 
er of 2, 2 is the fifth root of 32: &c. 

The extraction of the root is finding a number, which 
being multiplied into itself the requisite number of times, 
will reproduce the given number: for example, if we 
extract the square root of 81, we find it to be 9, because 
9X 9=81; but if we extract the biquadrate root of 81, 
‘we find it to be 3, because 3X 3X3XK3=—8l. 

Hence the root is designated by the number of times 
it is used _as factor in producing the corresponding pow- 
er. It is used twice in producing the second power, 
and is called the second root, or square root: it is used 
three times in producing the third power, and is called 
the third root, or cube root; it is used four times im pro- 
ducing the fourth power, and is called the fourth root, 
or biquadrate root: it is used five times in producing the 

fifth power, and is called the fifth root: &&« 

* A number, whose root can be exactly extracted, ts 
_ called a perfect power, and its root is called a rational’ 
number. For example, 4 is a perfect power of the 
second degree, and 2, its square root, is a rational num- 
ber; 27 is a perfect power of the third degree, and 3, its 
cube root, is a rational number; 64 is a perfect power of 
the second, third, and sixth degrees, and 8 its square 
roo, 4 its cube root, and 2 its sixth root, are rational 
numbers; = is a perfect power of the third degree, and 2, 
its cube root, is a rational number; .25 is a perfect power 
of the second degree, and .5, its square root, is a rational 
_ number. 

In short, any number, which is the exact root of any 
power, is a rational number, and its power a perfect 
14 


158 ARITHMETIC. XXVIIL. » 


power: and since any number may be the root of its cor- 
responding power, it follows that any root, which can be 
exactly expressed by figures, is a rational number. 

But there are numbers, whose roots can never be ex- 
actly extracted, and these numbers are called imperfect 
powers, and their roots are called irrational numbers, 
or surds. For example, 2 is not only an imperfect pow- 
er of the second degree, but an imperfect power of any 
degree, and not only its square root, but the root in eve- 


ry degree is irrational, or a surd; because no number, — 
either whole or fractional, can be found, which, being | 


involved to any degree, will produce 2. The same is 
true of many other numbers. In these cases, by using 


decimals, we can approximate, or come very near to the | 


root, which is sufficient for most purposes. Thus, we 
find the square root of 2 to be 1.414-++. The decimal 


may be carried to any number of places. 


Some numbers are perfect powers of one degree, and > 
imperfect powers of another degree. For example, 4_ 
is a perfect power of the second degree, and its square 
root, which is 2, is rational; but an imperfect power of | 


the third degree, and its cube root, which is 1.587 +, 
is a surd: 8 is an imperfect power of the second degree, 
and its square root, which is 2.828-++, is a surd; but 
a perfect power of the third degree, and its cube root, 
which is 2, is rational: 16 is a perfect power of the se- 
cond and fourth degrees, and its square root, which is 4, 
and its biquadrate root, which is 2, are both rational; but 
an imperfect power of the third degree, and its cube 
root, which is 2.519-+, is a surd. 

These irrational numbers or surds occur, whenever 
we endeavor to find a root of any number, which is not 
a perfect corresponding power; and, although they can- 
not be expressed by numbers either whole or fractional, 
they are nevertheless magnitudes, of which we may form 


an accurate idea. Jor however concealed the square 


root of 2, for example, may appear, we know, that it 
must be a number, which, when multiplied by itself, will 
exactly produce 2. This property is sufficient to give 
us an idea of the number, and we can approximate 
continually by the aid of decimals. 


XXVIII. EVOLUTION. 159 


A radical sign, written thus 4/, and read square root, 
is used to express the square root of any number, before 


which it is placed. The same sign with the index of the 


root written over it, is used to express the other roots: 
thus 4/ cube root: 4/ biquadrate root: ¥/ fifth root: &c. 
We will give the following radical expressions; 4/9 = 3 
A/ 8 = 2; 4/81=8; 4/32—2; these expressions are 
read thus; the square root of 9 is equal to 3; the cube 
root of 8 is equal to 2; the fourth root of 81 is equal to 
3; the fifth root of 32 is equal to 2. Hence it is evident 
that 46/9 XK 4/9=9; 4/8 X V/8 X 4/88; Kc. 

The explanation, which we have given of irrational 
numbers or surds, will readily enable us to apply to them 
the known methods of calculation. We know that the 
square root of 3 multiplied by itself must produce 38, 
which may be rales expressed, / 3 XK 4/ 3=3; also a/ 
BX BX V/8=3; VEXVI=HH V5 X 553 
ran aor ar? SSX A/S XK A/S XK 4/5 = 5; 


Instead of the radical sign, a fractional exponent is also 
used to express the roots of numbers. The numerator 
indicates the power of the number, and the denominator 


1 
the root. ‘Thus, 4° expresses the square root of 41, or 


4; 4°, the cube root of 4; 43, the biquadrate root of 4; 


1 2 
4°, the fifth root of 4; 8°, the cube root of the second 
power of 8; and since the second power of 8 is 64, and 


the cube root of 64 is 4, the expression 8? is equal to 4. 
The expression 48, is read thus, the square root of 
the third power of 4 is equal to 8. The expression 9° 
is equivalent to 4/9: and 8 is equivalent to 4/ 8°: also 
4} is equivalent to 4/47: the expression 4} is also equal 


to 4, because 7 is equal to 5. 
A fines or vinculum, drawn over several numbers, sig- 
nifies that the numbers under it are to be considered 


jointly: thus, 4/25-+-11 is equal to 6, because 25-+-11 is 


160. ARITHMETIC. XXIX. 


36, and the square root of 36 is 6; but 4/25 --11 is equal | 


to 16, because the square root of 25 is 5, and 5--11 ig 
16. The expression 4/27—6 -- 43 is equivalent to a/ 64 


And 4/100—73=8. Also 20—/9+-7+1=17. 
Likewise 4/90 —9 —4 + ./53—45 + 6= 13. 


X XIX. 
EXTRACTION OF THE SQUARE ROOT. 


-The product of a number multiplied by itself, is call- 


ed a square; and for this reason, the number, considered _ | 


im relation to such a product, is called a sQUARE ROOT. 


For example, when we multiply 12 by 12, the product, | 


144, is a square, and 12 is the square root of 144. 

If the root contains a decimal, the square will also_ 
contain a decimal of double the number of places; for 
example, 2.25 is the square of 1.5; and vice versa, if the 
square contains a'decimal, the square root will contain a 
decimal of half the number of places; for instance, 1.5 
is the square root of 2.25. 

In the upper line of the following table are arranged 
several square roots, and in the lower line, their squares. 


| Square roots. |1]2/8/4[5]6]718] 9110111112 ? 
| Squares. | 1/4] 9 1625.36.49 64/s1]100 121/144) 


When the square of a mixed number is required, it 
may be reduced to an improper fraction, then squared, 
and reduced back to.a mixed number. 

The squares of the numbers from 3 to 5, increasing 
by 4, are as follows. | 


Square roots. | 3| 33 | 34, hie [4 41 | 41 | 43 | 
| Squares. | 9 10-2, 1221141116] 18,1,| 204! 22%! 


J ———___ 


From this table we infer, that if a square contains a 
fraction, its square root also contains one; and vice versa. 


XXIXx. SQUARE ROOT. 161 


It is not possible to extract the square root of any 


_ number, which is not a perfect square; we can approxi 


Se 


mate the square root of such numbers, however, by the 
aid of decimals. 

When a root is composed of iwo or more factors, we 
may multiply the squares of the several factors together, 
and the product will be the square of the whole root; 
and conversely, if a square be composed of two or more 
factors, each of which is a square, we have only to mul- 
uply together the roots of those factors, to obtain the 
complete root of the whole square. For example, 2304 
= 4X 16x 36; the square roots of the factors are 2, 
4, and 6; and 2X4X6==48; and 48 is the square 
root of 2304, because 48 X 48=—= 2304. 

A square number cannot have more places of figures 
than double the places of the root; and, at least, but one 
less than double the places of the root. Take, for in- 
stance, a number consisting of any number of places, that 
shall be the greatest possible, of those places, as 99, the 
square of which is 9801, double the places of the root. 
Again, take a number consisting of any number of places, 


_ but Jet it be the least possible, of those places, as 10, 


the square of which is 100, one less than double the places 
of the root. 

As the places of figures in the square cannot be more 
than double the number of places in the root, whenever 
we would extract the square root of any number, we 
point it off into periods of two figures each, by placing a 


‘dot over the place of units, another over hundreds, &c. 


Thus 1936. The places in the root can never be more 
or less in number, than the number of periods thus 
pointed off. When the number of places in the given 
sum is an odd number, the left hand period will contain 


only one figure, as 169; but the root will nevertheless 
consist of as many places as there are periods; for 13 is 
the square root of 169. . 

The terms, square and square root, are derived from 
geometry, which teaches us that the area of a square is 
found by multiplying one of its sides by itself. 


62. ARITHMETIC. XXIX. 


The word arma signifies the quantity of space contained _ 
in any geometrical figure. te 


the length of aside of the an- | : i 
"i 


nexed square to be four feet, it i 


} 
| 
i 
i 


is evident that the figure contains ' | 
4 times 4small squares, each of | | 
which is 1 foot in length and 1 TAA 
foot in breadth; and since a foot | | 
in length anda foot in breadth ™ aa . 
constitute a square foot, the whole square contains 4_ 
times 4, or 16 square feet. If, instead of 4 feet, thal 
length of a side were 4 yards, the whole square would” 
contain 16 square yards; &c. Hence it is evident that — 
the area, which is 16, is found by multiplying a side of — 
the square by itself. v 
A PARALLELOGRAM is an oblong figure, 
having two of its sides equal and parallel to 
each other, but not of the same length with 
the other two, which are also equal and 
parallel to each other. We find the area, 
“or contents of a parallelogram by multiply- 
ing the length by the breadth: If we sup- 
pose the annexed right angled parallelogram 
to be 8 feet long.and 2 feet wide, it is 
manifest that it contains 2 times 8, or 16 
square feet; if the length were 8 yards and 
the breadth 2 yards, it would contain 16 
square yards; if 8 miles long and 2 miles 
wide, 16 square miles; &c. .We see that 
the area of this parallelogram is the same i 


with that of the preceding square; there- 
fore the square root of the area of a paral- 
lelogram gives the side of a square equal in 
area to the parallelogram. : 

It is further to be observed, that the square root of the — 
area of any geometrical figure whatever, is the side of a_ 
square, equal in area to the figure. ; 


. 


| 
XXIX. SQUARE ROOT. 163 


_ When the area of a square is given, the process of 
finding one of its sides, which is the root, 1s called the 
extraction of the square root, the principles of which we 
will now proceed to explain. | 
_ We have already learned, that a square number is a 
product resulting from two equal factors. For example, 
2025 is a square number resulting from the multiplication 
of 45 by 45. To investigate the constituent parts of this 
groduct, we will separate the root into two terms, thus, 
40--5, and multiply it by itself.im this form. We begin 
with multiplying 40-5 by 5, and set down the products 
separately, which are 200 4-25; we then multiply 40 +- 5 
xy 40, and set down the products separately, which are 
1600-+-200; the whole product, therefore, is 1600 -- 
200 + 200 + 25 = 2025; thus we see, that the whole 
wroduct or square contains the square of the first term, 
0 x 401600; twice the product of the two terms, 
0X 5 X 2400; and the square of the last term, 
1X 525. 

_ Now the extraction of the square root is the reverse 
if squaring or raising to the second power; therefore, 
he operation of extracting the square root of 2025, which 
ve know is the square of 45, must be performed in the 
nyerted order of raising 45 to the second power. 

We will now extract the square root of 2025, and 
‘xplain the process, step by step. | 
| | 2025 (45 

16 
a Divisor. 40X2=80 425 dividend 
Divisor, increased by last fig. 85 425 product of 85 by 5. 


Explanation of the precess. We began by separating 
he given number into periods of two figures each, putting 
i dot over the place of units, and another over hundreds, 
fe thereby ascertained that the root would contain two 
alaces of figures. We then found that the greatest 
square in the left hand period was 16, and placed its 
root, which is 4, in the quotient, and subtracted the 
square from the left hand period, and to the remainder 
osrought down the next period for a dividend. 


% 


' figure in the root must be such, that twice the product o 


' dend would have been divided by 10. Thus 8 is cor) 


164 ARITHMETIC. XXI¥ 


Then, knowing the figure in the root to be im the plac| 
af tens, and therefore equal to 40, and that the secon 


the first and second terms, together with the square ¢| 
the second, would complete the square, we took twic| 
the root already found, viz. 40 X 2== 80, for one of th 
factors, and using it for a divisor, found the second figut| 
in the root by dividing the dividend by this factor. 

Lastly, after finding that the second figure in the ro¢ 
was 5, we added it to the divisor, making 80-+ 588 
and multiplied the sum by 5, the last figure m theroo| 
and thus obtained twice the product of the two term,| 
and the square of the last term; because, 80 is twice th 
first term of the root, and being multiplied by 5, whic’ 
is the last term, the result is twice the product of th| 
two terms; and 5 being multiplied by 5, the product | 
the square of the last term. | 

It will be cbserved, that 4, the first figure in the roo 
being in the place of tens, was called 40, and double 
for a divisor; but, if we had merely doubled the roc 
without any regard to its place, making the divisor ¢ 
and had cut off the right hand figure of the dividend an 
divided what was left, the result would have been th 
same; because, in this operation, both divisor and div) 


tained 5 times in 42. The figure obtained for the roo! 
in this abridged method, would be placed at the right han| 
of the divisor, instead of being added thereto; thus, 88| 
making the completed divisor the same as before. Thi 
course, being the most concise, will be adopted in th! 
rule for extracting the square root, which we shall here 
after give. ) 


Suppose 169 square rods of land are to be laid out 1) 
a square, and the length of one of its sides is required. | 

We know that the length of a side must be such | 
aumber of rods, as, when multiplied by itself, will pro} 
duce 169; therefore we must extract the square root 0! 
the given number of square rods, and that root will b 
the answer. /169=13 Ans. 


XIX. SQUARE ROOT. 165 


, We here illustrate this 4 yo hay 
ast example geometrical- [7] 
, by a square figure A B 
_D, each side of which ¢}——-—- 
s 13 rods long. This 
jguare is divided into 169 
mall squares, each of 
vhich is a square rod. 

. The whole figure is also 
wubdivided into four fig- 
res, two of which, ef g 
aan hf iB, are squares; 
adthe other two, Ahfe ? Svinnk,.: 
jad C if g, are oblongs. 

, The square ef g Dis 10 rods on aside, and, therefore, | 
‘ontains 100 square rods. The oblongs are each 10 rods 
mg and 3 rods wide, and consequently each contains 30 
quare rods. The other square, B hfi, is 3 rods ona 
jide, and contains 9 square rods. 

, To illustrate the process of extracting the square root, 
ye shall take the side A B, which is divided into two 
jatts, the first of which, A h, is 10 rods long; the other, 
,B, 3 rods long. Ah being equal to ef, the square 
{A h, the first part, gives the area of the square D ef g; 
_B veing equal to h f, the area of the oblong A h fe, is 
ouna by multiplying the two parts, A h and h B, togeth- 
5 tne area of the other oblongifg C, is the same; there- 
re, the area of the two oblongs ts twice the product 
{the two parts, Ah andh B.. The square of the last 
art, h B, gives the area of the squaeh Bif. 

, We have therefore the square of the first part A h, 
0X 10—100 rods; twice the product of the two parts, 
hand h B, 10x 3X 2=60 rods; and the square of 
r last parth B, 3X 3=—9 rods. These being added 


»gether make 169 rods, the square of the whole figure — 

ABC D. : 

This illustration of a square corresponds exactly with 
jat of the first example, and of course the extraction of 
2 square root must proceed on the principles there 
xhibited. 


t 


166 ARITHMETIC. XXT 


From the illustrations of the two precedmg example 
we give the following rule for the extraction of the squa 
root. | é 

_ RULE. First—Point off the given number into perio 
of two figures each, by putting a dot over the place ¢ 
units, and another over every second figure to the le 
and also to the right, when there are decimals. — 

Secondly—Find the greatest square in the left has 
period, and write its root in the quotient. Subtract t 
square of this root from the left hand period, and to t 
remainder bring down the next period for a diwidend. 

Thirdly—Double the root already found, for a diviso 
Ascertain how many times the divisor is contatned ‘| 
the dividend, excepting the right hand figure, and pla 
the result in the root, and also at the right hand of t 
divisor. Multiply the divisor, thus increased, by t 
last fizure in the root, and subtract the product from ti 
dividend, and to the remainder bring down the ne 
period for a new dividend. | 
_ Fourthly—Double the root already found for a me 
divisor, and continue to operate as before, until all ti 


pertods are brought down. — | i 


It will sometimes happen, that, by dividing the divider 
as directed in the rule, the figure, obtained for the roo) 
will be too great. When this happens, take a less figur| 
and go through the operation again. 

When the places-in the decimal are not an even nun 
ber, they must be made so, by continuing the decime 
if it can be continued; if it cannot, by annexing a ciphe! 
that the periods may be full. r| 

If there be a remainder after all the periods are use 
a period of decimal ciphers may be added; or, if th! 

_ given number end ina decimal, the two figures that wou! 
arise from a continuation of the decimal. The operatic) 
may be thus continued to any degree of exactness. | 

If any dividend shall be found too small to contain th 
divisor, put a cipher in the root, and bring down the ne}| 
period to the right hand of the dividend for a new div’ 
dend, and proceed in the work. ¢ 


XXIX SQUARE ROOT. 167, 


_ When the square root of a mixed number is required, 
it will sometimes be necessary to reduce it to an improper 
fraction, or the vulgar fraction to a decimal, before ex- 
tracting the root. . 

_ If either the numerator or denominator of a vulgar frac- 
tion be not a square number, the fraction must be reduced 
to a decimal, and the approximate root extracted. 

i 
) 1. Extract the square root of 4579600. 


4579600(2140 Ans. 


~~ 


4 
( we siete so 
| Ist. divisor 4 57 first dividend. 
' 4] 41 
| Qd. divisor 42 1696 second dividend. 
oe 4.24 1696 
| 7 ae 00 


Tee. 
‘ 


_ 2. What is the square root of 110 33! 
a ’ 110.24(10.499-++ Ans. 
I 


ie 
ime, sist. divisor, 2 10 first dividend. 
. Qd. divisor, 20 1024 second dividend. 
= 204 = 816 
3d. divisor, 208 20824 third dividend. 
2089 ~—«18801 ‘ 
Ath. divisor, 2098 202324 fourth dividend. 


20989 188901 


- 13423 remainder. 

Reducing 24 to a decimal, we found it to be infinite, 
inthe recurrence of 24 continually; therefore, in con- 
tinuing the extraction of the root, instead of adding 
eriods of decimal ciphers, we added the period 24 each 
time. The extraction of the root might have been con- 
tinued indefinitely; but having obtained five places of 
figures in the root, we stopped, and marked off the three 
last places of the root for decimals; because we made 
use of three periods of decimals in the question. 


. What is the square root of 2704? 
. Extract the square root of 361. a| 
. What is the square root of 3025! i ‘) 
. What is the square root of 121? | 
. Extract the square root of 289. 
. Extract the square root of 400. 


. What is the square root of 848241? ‘| 
. Extract the square root of 3356224. © « | 
. What is the square root of 824464 ? 

. Find the square root of 49084036. 

. What is the square root of 688900? 

. Find the square root of 82864609. 

. Find the square root of 3684975616. 

. What is the square root of 44890000? 

. What is the square root of 165649? 

. Find the square root of 90484249636. 

. Find the square root of 26494625227849. 
. Find the square root of 262400.0625. 3 
. What is the square root of 841806.25 ? 4 
. What is the square root of 39:037504? ~7 || 
. Find the square root of 213.715161. 

. Find the square root of .66650896. i 
. What is the square root of 13340752? | 

. What is the square root of 152; ? 

. Extract the square root of 31834. 

. What is the square root of 51,4, ? 

. Extract the square root of 2%. 

. What is the square root of 55634? 

. Extract the square-root of 109674. 

. Find the square root of 4120900. 

. Extract the square root of 5. , 

. Extract the square root of 8. 

. Extract the square root of 84. 

. Extract the square root of 99. 

. Extract the square root of 101. 

. Extract the square root of 120. | 
. Extract the square root of 124. a 
. Extract the square root of 143.. 

. Extract the square. root of 1.5. a 


ARITHMETIC. XXD 


What is the square root of 4761? 


xix. | SQUARE ROOT. 169 


43. Extract the square root of .00032754. 
44. Extract-the square root of 2.3. 
45. Extract the square root of 2. 
46. Extract the square root of 3. 
47. Exxtract the square root of 2. 
| 48. Extract the syuare root of 32. 
49. Extract the square root of 1132. 
50. Extract the square root of 2673. 


The square root of the product of any two numbers is 
‘a mean proportional between those numbers. 

Thus, 41s a mean proportional between 2 and 8; be- 
cause 2:44:86. But when four numbers are propor 
tionals, the product of the extremes is equal to the pro- 
‘duct of the means;..that is, the product of the two given 
numbers is equal to the square of the mean proportional. 

51. Find a mean proportional between 4 and 256. 

52. Find a mean proportional between 4 and 196. 

53. Find a mean proportional between 2 and 12.5. . 

54. Find a mean proportional between 9.8 and 5. 

55. Find a mean proportional between 25 and 121. 

56. Finda mean proportional between 130 625 and 10. 
_ 57. Find a mean proportional between 52 and 545. 

* 58. Find a mean proportional between 2 and 34. 

59. Find a mean proportional between 12 and 147. 

60. Find a mean proportional between { and 4. 

61. Find a mean proportional between .5 and 98. 

_ 62. Find a mean proportional between 40623;%, 
and 828. 

63. Find a mean proportional between .25 and 1. 

64. Find a mean proportional between .1 and S10. 

65. Find a mean proportional between .04 and .36. 

66. Find a mean proportional between .09 and .49. 

67. Find a mean proportional between .2 and .018. 


When the square root of the product of the two givea 
numbers cannot be extracted without a remainder, the 
mean proportional is a’suRD, and may be approximated — 

by the aid of decimals. 

68. Find a mean proportional between 6 and 12. 

15 


_ table purpose; each man gave as many cents, as there 


- 


~ will it be ? 


170 - ARITHMETIC. XXIX. 


69. Find a mean proportional between 25 and 14. 
70. Find a mean proportional between 64 and 21. 
71.. Find a mean proportional between 46 and 55. 
72. Find a mean proportional between 5 and 81. 
73. Find a mean proportional between 77 and 19. 
74. A number of men spent 1 pound 7 shillings in | 
company, which was just as many pence for each man, | 
as there were men in the company. How many were 
there ° | 
75. A company of men made a contribution for a chari- 


were men in the company. The sum collected was 31 | 
dollars 36 cents. How many men did the company | 
consist of ? x 

76. If you would plant 729 trees in a square, how | 
many rows must you have, and how many trees in a row ? | 

77. A certain regiment consists of 625 men. How 
many must be placed in rank and file, to form the regi- 
ment into a square? 

78. It is required to Jay out 40 acres of land in a 
square. Of what length must a side of the square be? — 

79. It isrequired to lay out 20 acres of land in the 
form of a right angled parallelogram, which shall be twice | 
as long as it is wide. What will be its length and 
breadth? (See page 162.) 

80. It is required to lay out 30 acres of land in the 
form of a right angled parallelogram, the length of which 
shall be three times the width. How long and how wide 


A TRIANGLE is a figure 
having three sides and three 
angles. When one of the 
angles is such as would form 
one corner of a square, the 
figure is calleda right-angled 
triangle, and the following 
propositions belong to it. 


PROPOSITION ist. The square of the hypotennedin is 
equal to the sum of the squares of the other two sides. | 


| 
; 


| MXIX, SQUARE ROOT. © 171 


| PROPOSITION 2d. The square root of the sum of the 
squares of the base and perpendicular is equal to the 
hypotenuse. _ 
PROPOSITION 3d. The square root of the difference of 
the squares of the hypotenuse and base is equal to the 
| perpendicular. 
_ PROPOSITION 4th. The square root of the difference 
of the squares of the hypotenuse and perpendicular is 
equal to the base. 
_ By observing the above. propositions, when any two 
_ sides of a right-angled triangle are given, we may always 
_ find the remaining side. For example, suppose the base 
of the preceding figure to be 4 yards in length, and the 
perpendicular to be 3 yards in height; then the square 
of the base is 16 yards, and the square of the perpendicu- 
lar 9 yards, and the sum of their squares is 25 yards. 
The square root of 25 yards is 5 yards, which is the 
length of the hypotenuse. 

81. A certain castle, which is 45 feet high, is sur- 
rounded by a ditch, 60 feet broad. What must be the 
Jengih of a ladder, to reach from the outside of the ditch 
to the top of the castle ° 

82. A ladder 40 feet long, resting on the ground at 
the distance of 24 feet from the bottom ofa straight tree, 
and leaning against the tree, just reaches to the first limb. 
What is the iength of the tree’s trunk ? 

83. Two brothers left their father’s house, and went, 
one, 64 miles due west, the other, 48 miles due north, 
and purchased farms, on which they now live. How far 
from each other do they reside ” 

84. James and George, flying a kite, were desirous 
of knowing how high it was. After some consideration, 
they perceived, that their knowledge of the square root, 
and of the properties of a right angled triangle, would 

enable them to ascertain the height. James held the 

line close to the ground, and George ran forward till he 
came directly under the kite; then measuring the distance 
from James to George, they found it to be 312 feet; and 

pulling in the kite, they found the length of line out, to 
be 520 feet. How high was the kite ? 


ail 


XXIX. 


85. A ladder, 40 feet long, was so placed in a street, | 
as to reach a window 33 feet from the ground on one side, _ 
and when turned to the other side without changing the 
place of its foot, reached a window 21 feet high. The 
breadth of the street is required. ; 


86. The distance between the lower ends of two equal_ 
rafters, in the different sides of a roof, is 32 feet, and the — 
height of the ridge above the foot of the rafters is 12. 


feet. Find the length of a rafter. 


172 ARITHMETIC. 


A straight line, drawn [ 
through the centre of a square, Pa 
or through the centre of a right- 
angled parallelogram, from one 
angle to its opposite, is called’ 
a DIAGONAL; and this diagonal 
is the hypotenuse of both the 
right-angled triangles into w?.1ch 


the square or parallelogram is 
thus divided. i om NI 


87. A certain lot of Jand, lying in a square, contains | 
100 acres: at what distance from each other are the. 
opposite corners ? ; . 

88. There is a square field contaming 10 acres: what 
is the distance of the centre from either corner ? = 


bounded by one curve line, 
called the circumference, 
every part of which is equally 
distant from the centre. = 

A straight line through the B= 
centre of a circle is called a = 
- diameter, and a straight line 
from the centre of a circle to 


SS SSS { 

————————————————— = : 

the circumference is called a <= | 
radius. | pe eg | 
j 


¥ [ 
A cIRCLE is a plane surface 3 
; 


The areas of all circles. are to one another, as the 
squares of their like dimensions. That is, the area of a 
greater circle is to the area of a less circle, as the square __ 


| 


‘ XXIX. ~ SQUARE ROOT. 173 


, of the diameter of the greater to the square of the diame- 
ter of the less. Or thus, the area of the greater is to the 
| area of the less, as the square of the circumference of 

_ the greater to the square of the circumference of the less. 
) Therefore, to find a circle, which shall contain 2, 3, 
. 


4, &c. times more or less space than a given circle, we 


have the followng— 

RULE. Square one of the dimensions of the given 
| eircle, and, if the required circle be greater, multiply 
the square by the given ratio, then the square root of the 
| product will be the like dimension of the requ red circle; 
but, if the required circle be less than the given one, 
| dwide the square by the given ratio; then the square 
root of the quotient will be the similar dimension of the 
circle required. 

89. The diameter of a given circle is 11 inches: what 
is the diameter of a circle-containing 9 times as much 
space? 

90. Find the diameter of a circle, which shall, contain 
one fourth of the area of a circle of 42 feet diameter. 

91. What must be the circumference of a circular 
pond, to contain 4 times as much surface, as a pond, of 
43 mile in circumference ? 

) 92. Find the circumference of a pond which shall con- 
tain ; part as much surface, as a pond of 134 miles 
circumference. 

93. Find the diameter of a circle, which shall be 36 
times as much in area, as a circle of 184 rods diameter. 


The diameter of a circle is to the circumference in the 
ratio of 1-to 3.14159265, nearly: therefore, if we know 
_ the one, we can find the other. Thus, the circumference 
of a circle, the diameter of which is 8, is 3.14159265 X 
8—25.1327412; the diameter of a circle, the circum- 
ference of which is 15.70796325, is 15.70796325 + 3. 
14159265=5. 
To find the area of a cirlce, multiply the cireumference 
by the radius, and divide the product by 2. . 
94. How many feet in length is the side of a square, 
equal ih area to a circle of peel diameter ? 
. 15 


| * 


eo 


174 ARITHMETIC. XXX » 


95. Find the side of a square eaual in area to a circle | 
of 20 rods in diameter. a cl 
96. Find the diameter of a pond, that shall contain 4 | 
as much surface, as a pond of 6.986 miles circumference. | 
97. Find the length and breadth of a right-angled | 
parallelogram, which shall be 4 times as long as it is wide, | 
and equal in area to a circle of 43.9822971 rods circum- } 
ference. | 
98. Find the circumference of a pond, which shall | 
contain as much surface, as 9 ponds of ¢ of a mile diam-_ | 
eter each. | 


XXX. 


EXTRACTION OF THE CUBE ROOT, 


A CUBE isrepresented by a 
solid block—like either of me 
those annexed—with six plane | 
surfaces; having its length, 
breadth, and height all equal. 
Consequently, the solid con- 
tents of a cube are found by ; 
multiplymg one of its sides Aa 
twice into- itself. For this Oe 
reason, the third power of any_ jij 
number 1s called a cube. al 
_ Therefore, if we multiply 
the square of a number by its om 
root, we obtain a product, 
which is called a cube, or a 7 
cubic number. For instance, 4 multiplied by 4 produces 
16, which is the square of 4, as shown on one of the : 
sides of this larger block; and 16 multiplied by 4 pro- | 
duces ef, which is the cube of 4, as shown by the whole 
of the larger block. 4 
‘Thus the cube of any quantity is produced by multiply- A 
ing the quantity by itself, and again multiplymg the pro- — 
- duct by the original quantity. When the quantity to be — 


Dice 


XXX. CUBE ROOT. 175 


* 
cubed is a mixed number, it may be reduced to an im- 
proper fraction, and the fraction cubed, and then reduced 
back to a mixed number. 
_ As we can, in the manner explained, find the cube of 
a given number, so also, when a number is proposed, we 
‘may reciprocally find a number, which being cubed will 
produce the given number. In this case, the number 
sought is called, in relation to the given number, the 
‘cusE Root. ‘Therefore, the cube root of a given num- 
ber is the number, whose cube is equal to the given 
number. For instance, the cube root of 125 is 5; the 
cube root of 216 is 6; the cube root of } is 4; the cube 
root of 32 is 14. ; 
A cube cannot have more places of figures than triple 
the places of the root, and, at least, but two less ‘than 
triple the places of the root. - Take, for instance, a 
number consisting of any number of places, that shall be 
the greatest possible in those places, as 99, the cube of 
which is 970299; here the places are triple. Again, take 
a number, that shall be the least possible in those places, 
as 10, the cube of which is 1000; here the places are two 
less than triple. | . 
It is manifest from what has been said, that a cubic 
nuinber is a product resulting from three equal factors. 
For example, 3375 is a cubic number arising from 15 
15X15. To investigate the constituent parts of this cubic 
number, we will separate the root, from which !t was 
produced, into two parts, and instead of 15, write 10-5, 
and raise it to the third power in this form. 


10+ 5 

| 10 -+ 5 

Product of 10-+ 5 by 5, - - - - 50-+ 25 
Product of 10-++-5 by 10, - - 100+ 50 

The square, - - ae - 100+100 ++ 25 

eth Gt aa ae 

Prod. of 100-100-++25 by5, -  - 500+500--125 

Prod. of 100-+100-+25 by 10, 1000+1000+-250° 

he third power, - -  1000-+-1500-+'750-}- 125 


'e 
This product contains the cube of the first term, three 


ae | } 
ee ‘ - 
3 


% 


| wt a 
176 ARITHMETIC. XXX) 


times the square of the first term multiplied by the sec- 
ond term, three times the first term multiplied by the 
square of the second term, and the cube of the second 
term: thus, 10 10x 101000; 10X10X3X 5=1500; 
103 X 25 =750; 9X5 Xa = 125, 

Now, if the cube be given, viz. 1000-+ 1500-+-750- 
125, and we are required to find its root, we readily 
perceive by the first term 1000, what must be the first 
term of the root, since the babe root of 1000 is 10; if, 
therefore, we Sables the cube of 10, which is 1000, 
from the given cube, we shall have for a remainder, 
10X10X3X5 =1500, 10X3X25 =750, and 5X5X5 
‘==125; and from this remainder we must obtain the second 
term in the root. As we already know that the second 
term is 5, we have only to discover how it may be 
derived from the above remainder. Now that remainder 
may be expressed by two factors; thus, (10X10X3-—- 
10X3X5-+5xX5) 5: therefore, if we divide by three) 
times the square of the first term of the root, plus threé 
times the first term multiplied by the second term, plus 
the square of the second term, ‘the quotient will be the; 
second term of the root, which is 5.. 

But, as the second term of the root is supposed to te 
unknown, the divisor also is unknown; nevertheless we 
have the first term of the divisor, viz. three ‘times the 
square of the root already found; and by means of this; 
we can find the next term of the root, and then complete’ 
the divisor, before we perform the caviar After find- 
ing the second term of the root, it will be necessary, im 
order to complete the divisor, to add thrice the product | 
of the two terms of the root, and the square of the sec- 
ond term, to three times the square of the first term eS ' 
viously found. 

The preceding analysis explains the fotlevsas tule a 
the extraction of the cube root. 

RULE. First—Point off the given number into periods) | 
of three figures each, beginning al the wait’s place, and 
pointing lo the left in integers, and to the right in deci- 
mals; making full periods of decimals by supplying the. 
deficiency, when any exists. 

: ws 
“ 


4,5. 
1 ,) «+ 
1 ve 


(XXX. CUBE ROOT. 177 


| @dly—Find the root of the left hand period, place it 
‘im the quotient, and subtract its cube from the given 
‘number. The remainder is a new dividend. 

| 3dly— Square the root already found and multiply its 
square by 3, for a divisor. ; 
! Athly—Find how many times the divisor is contained 
in the dividend, and place the result in the quotient. 
| 5thly—mn order to complete the divisor, multiply the 
root previously found, by the number last put in the root, 

triple the product, and add the result to the divisor; also 
square the number last put in the root, and add tts square 
‘0 the divisor. 

Lastly—JMultiply the divisor thus completed, by the 
WiAber last put in the root, and subtract the product from 
the dividend. The remainder will be a new dividend. 

Thus proceed, till the whole root ts extracted. 


We will extract the cube root of 34965783, denoting 
sach step of the operation, from first to last, by a refer-- 
ance to that part of the rule, under which it falls. 


| BADGRTeS 
ddly. Cube of 300, subtracted, - - 27000000(300 
4 New dividend, - - - - - 7965783 
3dly. 3003003 {adivisor| 270000 : 
{thly. Divisorinnewdividend, - - - - = (20 
ithly. Triple prod.of 30020, 18000 
| Square of last number, 400 

Divisor completed, - 288400 
Lastly. 288400 X 20, subt’ed, - - | 5768000 
(3 New dividend, - - - - - 2197783 
3dly. 3203203 [adivisor] 307200 
tthly. Divisorinnewdividend, - - sive 
ithly. Triple prod. of 320X7, 6720 

Square of last number, 49 

Divisor completed, - 313969 
ia 313969 X 7, subtracted, | 2197783 


“ee © «© # ® 


300 -+-20-+-7 —327 Ans. 


178 ARITHMETIC. xXxy 


In completing every divisor, we have three produei 
to add together; viz. three times the square of the ro¢ 
already found; three times the product resulting from th 
multiplication of the reot already found, by the numbé 
Jast put in the root; and the square of the last number.*} 

If the ciphers be removed from the right hand of eae| 
of these products, the remaining figures in each succeed 
ing product will stand one place to the right of eac| 
preceding product; therefore, the work will be consider 
ably abridged by adopting the following— i 


RULE. First—Point off the given number into period} 
of three figures each, as before directed. ig 

2dly—Find the root of the left hand period, place 4| 
in the quotient without regard to local value, and sub| 
tract its cube from that period; and to the remainder bring| 
down the next period for a dividend. a | 

3dly—Square the root already found, without any re} 
gard to its local value, and multiply its square by 3, for 
a divisor. r] 

4thly—Find how many times the divisor is containes 
in the diwidend, omitting the two right hand figures, ane 
place the result in the quotient. 

Sthy— To complete the divisor, multiply the root pre 
viously found, by the figure last placed in the quotient. 
without regarding local value, triple the product, ana 
write u under the divisor, one place to the right; square 
the figure last put in the quotient, and write its square 
under the preceding product, one place to the right. Add 


these three together, and their sum is the divisor comp lela 


| 


| 
| 


% 


Lastly—.Multiply the divisor thus completed, by the 
figure last placed in the root, and subtract the produgt 
from the dividend; and to the remainder bring down the 
next period for a new dividend. i 

Thus proceed, till the whole root is extracted. + 


Observe, that, when the divisor is not contained in the 
dividend, as sought in the fourth part of the rule, a cipher 
must be put in the root, and the next period brought down 
for anew dividend. 


Shi) 
e a sa nx 
Pe 
> ‘ 


iy ; 


| 


ix_xx. CUBE ROOT. 179 


Observe, also, that when the figure obtained for the 
‘root by dividing, as directed in the fourth part of the rule, 
jis found, on completing the divisor, to be too large, a 
ssmaller figure must be substituted in its place, and the 
divisor completed anew. 

_ There are always as many decimals in the root, as 
jperiods of decimals in the power. 

£ We will extract the cube root of 6589031 1319, in the 
abridged form; referring, as before, to the particular part 
of the rule, under which each step of the operation pro 
ceeds. 


First, 65890311319(4039 
Qdly. Cubeof4,subt’d - - - (64 
} Dividend, - - - - - 1890 
Sdly. 4x4~x8[divr] - - 48 
Athly. 48 was not con- 

ad tamed in 18. - - - | 0000 
Lastly. New dividend, - - - | 1890311 
Sdly. 40x40x3, - - 4800 
4thly. 4800 in 18903, 
leg 3 times. 
dthly. Triple product 

i GE4A0 Sars Ob 

Square of 8, - - - 9 


ie Divisor comp’d, - 483609 

Lastly. 4836093, and 

subtracted, - - - - -| 1450827 
New dividend, - - - 439484319 
Sdly. 4034033, - .487227 

4thly. 487227 in 4394 


843, 9 times. 
Sthly. Triple product eae 
of 403x9, - 10881 
Square of 9 - 81 


Divisorcomp’d, 48831591 
astly. 48331591 x 9, 
, andsubtracted, - - - 439484319 


- Ans. 4089 


180 ARITHMETIC. 


without reference to the parts of the rule. 


178263.433152(56.28 Ans. | 


125 
: 75° | 53263 
90 
36 
~ 8436) 50616 
0408 | 2647433 
300 ; 
Ro Bucs, 
944164) 1888328 
947532 759105152 
13488 
64 


94888144} 759105152 


oie ie Wm 6 8 (ee ie: 


. Extract the cube root of 614125. 


i 

2. What is the cube root of 191102976 ? 

3. What is the cube root of 18399.744 ? 

4. Find the cube root of 253895799552. 

5. What is the cube root of 1740992427 ?- 
6. Extract the cube root of 35655654571. 
7. Find the cube root of 27243729729. 

8. What is the cube root of 912673000000 ? 
9. What is the cube root of 67518581248 ? 


XxX) 


We will now extract the cube root of 178263.43315: 
m the avridged form, as in the preceding example; | 


. Find the cube root of 729170113230343. 
. Extract the cube root of 643.853447875 


Find the cube root of .000000148877. 


. What is the cube root of 123? 

. Extract the cube root of 517. 

. Extract the cube root of 900. 

. Extract the cube root of 74 
. What is the cube root of $ : r 

. What is the cube root of 2 37° 

. What is the cube root of 725% ? 

. Extract the cube root of 26. 


| 


i 
| 


% . ¢ 
aoe ae —s = — nein — LS STI EA ~_ cs = om = 


\RXX. CUBE ROOT. isl 


| 
i To find two MEAN PROPORTIONALS between two given 
mumbers, divide the greater by the less, and extract the 
cube root of the quotient: then multiply the sube root by 
| the least of the given numbers, and the product will be 
the least of the mean proportionals; and the least mean 

proportional multiplied by the same root, will give the 
greatest mean proportional. 

21. What are-the two mean proportionals between 6 

and 750? 
_. 22. What are the two mean proportionals between 56 
and 12096? 


To find the side of a cube equal in solidity to any 
given solid, extract the cube root of the solid contents of 
the given body, and it wiil be the required side. 

23. There is a stone, of cubic form, containing 21952 

solid feet. What is the length of one of its sides? 

24. The solid contents of a globe are 15625 cubic 
inches: required the side of a cube of equal solidity. 

25. Required the side of a cubical pile of wood, equal 

to a pile 28 feet long, 18 ft. broad, and 4 ft. high. 


. 


All solid bodies are to each other, as the cubes of their 
‘diameters, or similar sides. 

26. If a ball 6 inches in diameter weighs 32 pounds. 
what is the diameter of another ball of the same metal, 
weighing 4 pounds ? 

27. If a ball of 4 inches diameter weighs 9 pounds, 
what is the diameter of a ball weighing 72 pounds ? 

28. What must the side of a cubic pile of wood mea- 
sure, to contain 4 part as much as another cubic pile, 

_ which measures 10 feet ona side? 

29. If 8 cubic piles of wood, each measuring 8 feet on 
a side, were all put into. one cubic pile, what would be 
the dimensions of one of its sides ? 

30. The solid contents of a globe 21 inches in diame- 
ter are 4849.0596 solid inches; what is the diameter of 
a globe, whose solid contents are 11494.0672 inches ? 

31. What are the inside dimensions of a cubical bin, 
that will hold 85 bushels of gram? (See note, page 27.) 
32. What must be the inside dimensions of a cubical 

16 


——— 


182 ARITHMETIC. XXXII. 


bin, to hold 450 bushels of potatoes, 2815.489 cubic 


inches, (heaped measure), making a bushel ? 


33. What must be the inside measure of a cubical | 


cistern, to hold 10 hogsheads of water ? 
34. What must be the inside measure of a cubical 
cistern, that will hold 20 hogsheads of water ? 


| 


| 


35. What are the inside dimensions of a cubical cis- 


tern, that holds 40 hogsheads of water? 


36. Suppose a chest, whose length is 4 feet 7 inches, - 
breadth 2 feet 3 inches, and depth 1 foot 9 inches: what 


is the side of a cube of equal capacity ? 

37. Suppose I would make a cubical bin of sufficient 
capacity to contain 108 bushels; what must be the dimen- 
sions of the sides ? : 


» 


XXXI. 
ROOTS OF ALL POWERS. 


The roots of many of the higher powers may be ex- 
tracted by repeated extractions of the square root, or 
cube root, or both, as the given power may require. 


Whenever the index of the given power can be resolved 


into factors, these factors denote the roots, which, being 
successively extracted, will give the required root. 


Thus, the index of the fourth power is 4, the factors” 


of which are 2X2; therefore, extract the square root 
of the fourth power, and then the square root of that 
square root will be the fourth root. The sixth root is 
the cube root of the square root, or the square root of 
the cube root; because 3 X2—6. The eighth root is 
the square root of the square root of the square root; be- 
cause 2X2 2=—8. The ninth root is the cube root 
of the cube root; because 3X 3—9. The tenth root 
is the fifth root of the square root; because 2K 5=10. 
The twelfth root is the cube root of the square root of 
the square root; because 2X 2X*3—12. The twenty- 
seventh root is the cube root of the cube root of the cube 
TOOL; BDECAUSE SX ho 0 faa 


| XXXII ROOTS OF ALL POWERS. 183. 


The following is a GENERAL RULE for extracting the 


roots of all powers. 


RULE. First—Prepare the givennumber for extraction, 


by pointing off from the unit’s place, as the required root 


directs; that is, for the fourth root, into periods of four 


figures; for the fifth root into periods of five figures, §c. 


2dly.—Find the first figure of the root by trial, and 


_ subtract its power from the left hand period. 


3dly.— To the remainder bring down the first figure in 


the next period for a dividend. 


4thly.—Involve the root to the next inferior power to 


that which is given, and multiply it by the number denot- 
ing the given power, for a divisor. 


5thly.—Find how many times the divisor ts contained 
in the dividend, and the quotient will be another figure 
of the root. | 

6thly.— Involve the whole root to the given power, and 
subtract it from the two left hand periods of the given 
number. 

Lastly.— Bring down the first jigure of the next period 
to the remainder, for a new dividend, to which find a new 
divisor, as before. Thus proceed, till the whole root ts 
extracted. 

Observe, that when a figure obtained for the root by 
dividing, is found by involving, to be too great, a less 
figure must be taken, and the involution performed again. 


We will extract the fifth root of 36936242722357. 


36936242722357(517 Ans. 
55—. 3125 | 
54 & 5 = 3125 first divisor. 5686 first dividend. 
Rit S45085251 
Ben <5 ==33826005, 
second divisor. 943371762 2d. dividend. 
YW ie 369362427 22357: 
. What is the fifth root of 5584059449 
. Find the fifth root of 2196527536224. 


. Extract the fifth root of 16850581551 ° 
. Find the seventh root of 2423162679857794647 


- - 


m C9 to = 


184 ARITHMETIC. XXXIV) 


XXXII. 
EQUIDIFFERENT SERIES. 


Aseries of numbers composed of any number of terms, 
which uniformly increase or decrease by the same num- | 
ber, 1s called aw EQUIDIFFERENT SERIES. ‘This series 
has, very commonly, but without any propriety, been | 
called Arithmetical Progression. | 

When the numbers increase, they form an ascending — 
series; but when they decrease, a descending series. | 
Thus, the natural numbers, 1, 2, 3, 4, 5, 6,7, 8,9, form | 
an ascending series, because they continually increase by _ 
1; but 9, 8, 7, 6, &c. form a descending series, because | 


they continually decrease by 1. | 

‘he numbers, which form the series, are called the 
terms of the series. The first and last terms in the series _ 
are called the extremes; and the other terms, the means. 

Lhe number, by which the terms of the series are 
continually increased or diminished, is called the common 
difference. Therefore, when the first term and comu:ion 
difference are given, the series may be continued to any — 
length. For imstance, let 1 be the first term in an equi- 
different series, and 3 the common difference, and we 
shall have the following increasing series; 1, 4, 7, 10,) 
13, &c., in which each succeeding term is’ found by 
adding the common difference to the preceding term. 

THEOREM I. When four numbers form an equidiffer- 
ent serves, the sum of the two extremes is equal to the sum 
of the two means. Thus, 1, 3, 5, 7, is an equidifferent 
series, and 1+7==3-+-5. Also in the series 11, 8, 5, 
2. To Be, 

THEOREM Il. Jn any equidifferent series, the sum of 
the two extremes is equal to the sum of any two means, 
that are equally distant from the extremes; and equal to 
double the middle term, when there is an uneven number 
of terms. Take, for example, the equidifferent series, 
2, 4, 6, 8, 10, 12, 14;2+14—=4+412; and2+14= 
6+10; also 2+14=—8-+18. 


XXXII. EQUIDIFFERENT SERIES. 185 


Since, from the nature of an equidifferent series, the 
second term is just as much greater or less than the first, 
as the last but one is less or greater than the last, it is 
evident, that when these two means are added together, 
the excess of the one will make good the deficiency of 
the other, and their sum will be the same with that of 


_ the two extremes. In the same manner it appears, that 


the sum of any other two means equally distant from the 

extremes, must be equal to the sum of the extremes. 
THEOREM Ill. The difference between the extreme 

terms of an equidifferent series is equal to the common 


difference multiplied by the number of terms less 1. 
_ Thus, of the six terms, 2, 5, 8, 11, 14, 17, the common 
_ difference is 3, and the number of terms less 1, is 5; then 
_ the difference of the extremes is 17—2, and the common 


difference multiplied by the number of terms less 1, is 


i Xx 5; and 17—2=35. 


The difference between the first and last terms, is the 
increase or diminution of the first by all the additions or 
subtractions, till it becomes equal to the last term: and, 
as the number of these equal additions or subtractions is 
one less than the number of terms, it is evident that this 
common difference being multiplied by the number of 
terms less 1, must give the difference of the extremes. 

THEOREM Iv. The sum of all the terms of any equi- 


different series is equal to the sum of the two extremes 


4 
ft 


multiplied by the number of terms and divided by 2; or, 
which amounts to the same, the sum of all the terms is equal 
to the sum of the extremes multiplied by half the thenum- 
ber of terms. For example, the sum of the following 
geties, 2, 4, 6, 8, 10, 12, 14, 16, is 2116 X4=-72. 
This is made evident by writing under the given series 
the same series inverted, and adding the corresponding 
terms together as follows. 


The given series, 2, 4, 6,°°,10, 12 14,16. 
Same series inverted, 16,14,12,10, 8, 6, 4, 2. 
Sums of the series, 13,15,18, (8. tsoioe15,18 


This series of equal terms, (18), is evidently equal tc 
twice the sum of the given series; but the sum of these 


16% 


| 
IgG ARITHMETIC. XXXII | 


LY 


equal terms is 18 X 8== 144; and since this sum is twice: | 
as great as that of the given series, the sum of the given 
series must be 72. 
| 
Any three of the five following things being given, the | 
other two may be readily found. 
The first term. 
The last term. 
The number of terms. | 
The common difference. 
The sum of all the terms. 


PROBLEM. I. The extremes and number of terms be- 
ing given, to find the sum of all the terms. } 
RULE. Multiply the sum of the extremes by the num-_ 
ber of the terms, and half the product will be the sum | 
of all the terms. See Theorem 4th. | 
1. The first term in an equidifferent series, 1s 3, the last 
term 19, and the number of terms is 9. What is the | 
sum of the whole series ? 
2. How many strokes does a common clock strike in 
12 hours ? ) | 
3. A hundred cents were placed in a right line, a yard | 
apart, and the first a yard from a basket. What distance | 


did the boy travel, who, starting from the basket, picked 
them up singly, and returned with them one by one to | 
the basket? q 
4. If a number of dollars were laid in a straight line | 
for the space of a mile, a yard distant from each other, _ 
and the first a yard from a chest, what distance would | 
the man travel, who, starting from the chest, should pick | 
them up singl~, returning with them one by one to the | 
chest ? , 3! 
PROBLEM Il. The extremes and number of terms — 
given, to find the common difference. | 
RULE. Subtract the less extreme from the greater, and | 
divide the remainder by the number of terms less 1, and 
the quotient will be the common difference. | 
It has been shown under Theorem 3d. thatthe differ- 


t 
: 


4 » 
XXXIT = EQUIDIFFERENT SERIES. 187 


| ence of the extremes is found by multiplying the common 
difference by the number of the terms less 1; conse- 
quently, the common difference is found by dividing the 

Biference of the extremes by the number of the terms. 
less 1. 

5. A man had 10 sons, whose ages differed alike; the 

youngest was 2 years old, and the eldest 29. What 
,was the difference of their ages? - 
| 6. The extremes in an equidifferent series are 3 and 
$7, and the number of terms 43. Required the common 
difference. 
__ 7. A man is to travel from Boston to a certain place 
in 9 days, and to go but 5 miles the first day, and tc 
‘Increase his journey every day alike, so that the last day’s 
journey may be 37 miles. Required the daily increase, 
and also the number of miles travelled. 


_ PROBLEM IN. The extremes and common difference 
given, to find the number of terms. 

RULE. Divide the difference of the extremes by the 
common difference, and add 1 to the quotient; the sum 
will be the number of terms. ‘ 

The difference of the extremes divided by the number 
of the terms less 1, gives the common difference; con- 
‘sequently, the same divided by the common difference 
must give the number of terms less 1: hence, this quotient 
augmented by 1, must give the number of terms. 

8. The extremes in an equidifferent series are 3 and 
39, and the common difference is 2: what is the number 
-of terms? 

_ 9. A man going a journey, travelled 7 miles the first 
day, and increased his journey every day by 4 miles, and 
the last day’s journey was 51 miles. How many days did 
he travel, and how far ? 

*10. A mancommenced a journey with great animation, 
and travelled 55 miles the first day; but on the second 
day he began to be weary, and travelled only 51 miles, 
‘and thus continued to lose 4 miles a day, till his las: 
day’s journey was only 15 miles. How many days did 
he travel ? 


‘| 
| 
‘| 
| 


| 


188 ARITHMETIC. XXXII. 


PROBLEM IV. 'To find an equidifferent mean between 
two given terms. 

RULE. Add the two given terms together, and half 
their swin will be the equidifferent mean required. | 

11. Find an equidifferent mean between 3 and 15. 

12. What is the equidifferent mean between 7 and 53? 

13. Find an equidifferent mean between 5 and 18. 


i 
PROBLEM V. ‘To find two equidifferent means between | 
the given extremes. } | 
RULE. Dwwide the difference of the extremes by 3, and. 
the quotient will be the common difference, which, being 
continually added to the less extreme, or subtracted, from | 
the greater, gives the two required means. | ij 
14, Find two equidifferent means between 4 and 13. ° 
15. Find two equidifferent means between 5 and 22. | 
16. Find two equidifferent means between 4 and 53. | 
PROBLEM VI. ‘To find any numbers of equidifferent 

means between the given extremes. | 
RULE. Divide the difference of the extremes by the re- 
quired number of means plus 1, and the quotient will be 
the common difference, which being continually added to | 
the less extreme, or subtracted from the greater, will give | 
| 


the mean terms required. 
17. Find five equidifferent means between 4 and 28. 
18. Find six equidifferent means between 6 and 55. _ | 
19. Find 3 equidifferent means between 34 and 142. 
29. Find one equidifferent mean between 56 and 100. 


XXXII. ' 7 
CONTINUAL PROPORTIONALS. 


The numbers of a series in which the successive terms: 
increase by a common multiplier, or decrease by a com- 
mon divisor, are CONTINUAL PROPORTIONALS. : 

This series of numbers has been commonly called a 
Geometrical Progression; but, perceiving no appropriate 


| 

| 

; 

XXXHT. CONTINUAL PROPORTIONALS. 189 


meaning in this term, we choose to call the series, what 
‘itis in truth, a series of Continual Proportionals. 
__ The common multiplier, or common divisor, by which: 
the successive terms are increased or deminished, is 
ealled the ratte of the series, or the common ratio. 

Ue 2 3 4) 5 6 ’ 
| Thus, 3, 6, 12, 24, 48, 96 is a series of continual pro- 
portionals, in which each successive term is produced by 
multiplying the preceding term by 2, which is the com- 
lion ratio. The numbers 1, 2, 3, 4, &c. standing 
above the series, mark the place, which each term holds — 
in the series. 
Also, 729, 243, 81, 27,9, 3, 1, is a series of continual 
proportionals, in which each successive term is found by 
dividing the preceding term by 3, the common ratio. 

In an increasing series, the ratio is the quotient, which 
results from the division of the consequent by the ante- 
cedent; but in a decreasing series, the ratio is the quo- 
tient resulting from the division of the antecedent by the 
consequent. 

In every series of continual proportionals, any four 
successive terms constitute a proportion. Thus, in the first 
of the above series, 3: 6=12: 24, and 6: 1224: 
48, also, 12: 24—48:96. In the second series, 
Oey eto 3): 27, 2435 81 Rte Oy oli: QI 9 
3, 27: 9==3:1. Therefore, when there are only four 
erms, the product of the extremes is equal to the pro- 
duct of the means. 

Furthermore, in any series of continual proportionals, 
he product of the extremes is equal to the product of 
my two terms equally distant from them; and equal to 
he second power of the middle term, when there is an 
meven number of terms. For instance, take the con- 
inual proportionals 2, 4, 8, 16, 32, 64, 128; then 2x 
'28—=4 X 64; also 2X 128=8 X 32; and 2K 128=— 
16 X 16. 

When the first term and the ratio are given, a series 
pf continual proportionals may be extended to any num 
ver of terms by continually multiplying by the ratio ir 
mM increasing series, or dividing in a decreasing series 


190 ARITHMETIC. XXXII 


For example, the first term being 2, and the ratio 3, if we 
make it an increasing series by continually multiplying by 
the ratio, we obtain the following series, 2, 6, 18, 54, 
162, 486, which may be extended to any number of 
terms; but, if we make it a decreasing series by continu! 
ally dividing by the ratio, we obtain the following series, 
2, 3) 3) He) Siro a¥z) Which may also be extended to any 
number of terms. 

In the series 2, 6, 18, 54, 162, 486, we obtain the| 
second term by multiplying the first term by the ratio; 
the third term by multiplying the second term by the 
ratio; the fourth term by multiplying the third term by 
the ratio; the fifth term by multiplying the fourth term by 


the first term by the fifth power of the ratio. The fifth 
power of 3 is 243, which being multiplied into the first 
term, the result is 486, the same as in the series. 4 
Hence we see, that any term in any increasing series 
of continual proportionals may be found by multiply iti 
the first term by that power of the ratio, which is denoted. 
by the number of terms preceding the required one.: 
For instance, the ninth term in an increasing series is 
found by multiplying the first term by the eighth power | 
of the ratio; thus let 2 be the first term, and 3 the com>: 
mon ratio; then 235 gives the ninth term, which is| 
13122. . a 
If the series be a decreasing one, any term in it may 
be found py dividing the first term by that power of the 
ratio; which is denoted by the number of terms preceding. 
the required one. For instance, the seventh term in a 
decreasing series is found by dividing the first term by 
the sixth power of the ratio; thus, let 24576 be the first. 
term in a decreasing series, and 4 the common ratio;| 
then 24576 + 4° gives the seventh term, which is 6. a 
We will now state several problems, which occur iP 
Sa hig proportionals, and give the rules for performing: 
them. | | 


—— aus 


: XXXIIl. CONTINUAL PROPORTIONALS. 191 


PROBLEM I. ‘The first term and the ratio being given, 
to find any other proposed term. 
RULE. Raise the ratio to a power, whose index is equal 
to the number of terms preceding the required term: then, 
uf it be an increasing series, multiply the first term by 
this power of the ratio; but, if it be a decreasing series, 
divide the first term by it: the result will be the required 
term. | 

1. Required the eighth term in an increasing series 
whose first term is 6, and ratio 2. 
* 2. Required the ninth term in a decreasing series, the 
‘first term of which is 131072, and the ratio 4. 
_ 8. What is the seventh term in an increasing series, 
ithe first term being 3, and the ratio 1.5? 
__ 4, What is the sixth term in an increasing series, whose 
first term is z¢357, and ratio 7 ? 
_ 5. What is the tenth term in a decreasing series, the 
first term being 387420489, and the ratio 9? 


| 
i 
| 
| 
} 


One of the principal questions, which occurs in a series 
of continual proportionals, is to find the sum of the 
series. We shall, therefore, illustrate the method. 

Let there be given the following series, consisting of 
seven terms, whose common ratio is 3; viz. 2, 6, 18, 54, 
162,486, 1458. Let each term in this series be multiplied 
by the ratio 3; and let each product be removed one 
place to the right of the terms in the given series. 

The given series, 2,6, 18, 54, 162, 486, 1458 
multiplied by ratio. 6, 18, 54, 162, 486, 1458, 4374. 

Now the last term in the second series is produced by 
multiplying the last term in the given series by the ratio; 
and it is evident that if the given series be subtracted 
from the second series, the remainder will be the last 
erm in the second series diminished only by the first 
‘erm in the given series, and this remainder will be twice 
the sum of the given series; consequently, if we divide 
t by 2, the quotient will be the sum of the given series 
out 2 is the ratio less 1. Hence 
_ PROBLEM II. The extremes and the ratio being given, 


‘o find the sum of the series. 
e 


192 ARITHMETIC. XXXI0 


RULE. Multiply the greater extreme by the ratio, fron 
the product subtract the less extreme, und divide the re 
mainder by the ratio less 1, and the quotient will be th 
sum of the series. 

6. The first term in a series of continual proportional) 
is 1, the last term is 65611, ana the ratio is 3. What 1) 
the sum of the series ? 

wae 65611 < 3= 196833 
| 1 
ratio 3—1=2)196832 

98416 Ans. 

7. The extremes of a series of continual proportional 
are 3 and 12288, and the ratio is 4. What is the sun 
of the series ° 

8. The first term in a series of continual proportional 
in 12500, the last term is 4, and the ratio 5. What i) 
Hy sum ef the series? | 

. The first term in a series of continual proportional 
Me the last term 1792, and the ratio is 2. What is the 
sum of the series? 

10. The extremes ima series of continual proportional 
are 5 and 37.96875, and the ratio 1s 1.5. _ What is the 
sum of the series ? f 

11. The first term ina series of contmual proportional: 
is 100, the last term .O1, and the ratio 2.5. Requing 
the sum of the series. | 


7 a 


| 
| 
‘| 
| 


| 


PROBLEM I. ‘The first term, the ratio, and the num: 
ber of terms given, to find the sum of the series. ! 
RULE. Find the last term by problem 1, and the su 
of the series by problem 2. | 
12. The first term in an increasing series of continua} 
proportionals i is 6, the ratio 4, and the number of terms 8. 
What is-the sum of the series ? | 
13. The first jenn in an increasing series of f contiaill 
proportionals is 1, the ratio 4, and the number of terms 
13. Required the sum of the series. . 
14. The first term in a decreasing series of contin 
proportionals i is 1, the ratio 3, and the number of term 
12. What is the sum of the series? “ 


-——_— 


XXXII. CONTINUAL PROPORTIONALS. 193 


15. A man offers to sell his horse by the nails in his 
shoes, which are 32 in number. He demands one mill 


. for the first nail, 2 for the second, 4 for the third, and so 


on, demanding for each nail twice the price of the pre-’ 


/ ceding. It is required to find what would be the price 


of the horse. 

16. An ignorant fop wanted to purchase an elegant 
house, and a facetious man told him he had one, which 
he would sell him on these moderate terms; viz. that he 


, should give him one cent for the first door, 2 for the 


—S 


_ second, 4 for the third, and so on, doubling the price for 
| every door, there being 36. It is a bargain, cried the 


simpleton, and here is a half-eagle to bind it. What was 
the price of the house? 

PROBLEM IV. The extremes and the number of terms 
being given, to find the ratio. 

RULE. Divide the greater extreme by the less, and the 
quotient will be that power of the ratio, which is denoted 
by the number of terms less 1; consequently, the corres- 
ponding root of this quotient will be the ratio. 

This problem is the reverse of problem 1, and the 
reasoning which precedes that problem, ae netnay eluci- 
_ dates the rule in this. 

The first term im a series of proportionals is 192, the 
Jast term 3, and the number of terms 7. What is the ratio? 

192-3—64; the number of terms less 1, is 6; there- 
| fore the sixth root of 64, which is 2, 1s the ratio. 

17. Ina series of continual proportionals, the first 
/term is 7, the last 45927, and the number of terms, 9. 
_ What is the ratio? | 
| 18. The first term in a series of continual proportion- 
als is 26244, the last term 4, and the number of terms 5. 

Required the ratio. 

19. The first term ina series of continual proportionals 
is 1, the last term 1029427, and the number of terms 8. 
, What is the ratio? | ; 
| 20. The first term ina series of continual proportion- 
als is 78125, the last term z45, and the number of terms 

11. Required the ratio. 


17 


194 ARITHMETIC. XXXL! 


PROBLEM V. ‘To find any number of mean proportion-_ 
als between two given numbers. ; 
RULE. The two given numbers are the extremes of a 
series consisting of two more terms than there are means 
required; hence the ratio will be found by problem 4.) 
Then the product of the ratio and the less extreme will 
be one of the means; the product of this mean and the 
ratio will be another mean; and so on, till all the required 
means are found. ' 
When only one mean is required, it is the square root} 
of the product of the extremes. . | 
21. Find 3 mean proportionals between 5 and 1280. | 
Here the series is to consist of five terms, and the) 
extremes are 5 and 1280; hence the ratio is found by the’ 
fourth problem to be 4; and by the repeated multiplica=| 
tion of the least term by the ratio, the means are found 
to be 20, 80, and 320. . 
22. Find four mean proportionals between 4 and 2401. 
23. Find five mean proportionals between the num- 
bers, 279936 and 6. 
24. Find a mean proportional between 1 and 2809. 


COMPOUND INTEREST BY SERIES. 


It has been shown in ArT. xv, page 107, that cora- 
pound interest is that which arises from adding the interest 
to the principal at the end of each year, and taking the | 
amount for a new principal. Now, the several amounts! 
for the several years form a series of continual propor- 
tionals; and, to find the amount for any number of years, 
we may adopt the following— 

RULE. Find the last term of an increasing series of 
continual proportionals, whose first term is the principal, 
whose ratio is the amount of 1 dollar for 1 year, and whose’ 
number of terms is the number of years plus1. The 
last term is the required amount. See Problem Ist. 

In the examples under this rule, no more than six deci- _ 
mal places need be included. ] 

25. What is the amount of $100, at 6 per cent. | 
compound interest, for 4 years ? | 


f 


) 
pi, 


XXXII CONTINUAL PROPORTIONALS. 195 


J 


) 26. What is the amount of ‘75, at 5 per cent. com- 
_ pound interest, for 9 years? 

27. What is the amount of §294, at 4 per cent. com- 
pound interest, for 7 years? 
28. Find the amount of $18.25, at 7 per cent. 
“compound interest, for 12 years. 

29. Find the amount of $751.30, at 5 per cent. 
compound interest, for 8 years. 

30. Find the arhount of $4798, at 6 per cent. com- 
pound interest, for 12 years. 

31. What is the amount of $5.14, at 7 per cent. 
compound interest, for 16 years? What is the interest ? 

32. What is the compound interest of $1000 for 20 
years, at 6 per cent.? What is the amount? 


+ 


COMPOUND DISCOUNT. 


Discount corresponding to simple interest has already 
been treated, in Art. xvi; but discount corresponding 
to compound interest is now to be computed. 

On the supposition that money can be let out at com- 
pound interest, the present worth of a debt, payable at a 
future period without interest, is that principal, which, at 
compound interest, would give an amount equal to the 
debt, at the period when the debt is payable. 

RULE. Find the last term of a decreasing series of 
continual proportionals, whose first term is the debt, whose 
ratio is the amount of 1 dollar for 1 year, and whose 
number of terins is the number of years plus 1. The last 
term is the present worth. See Problem Ist. 

33. What principal, at 10 per cent. compound interest, 
will amount, in 4 years, to $8.7846? 

34. What is the present worth of $68.40, payable 11 
years hence; allowmg discount according to 5 per cent. 
compound interest ? 

35. What is the present worth of $350, ls in 5 
years; allowing discount at the rate of 6 per cent. com- 
pound interest ! > 

36. What is the present worth of $3525, ae in 3 
years; discount being allowed as in the last example ? 


196 ARITHMETIC. XXXIV 


37. How much must be advanced to discharge a debt 
of $700, due in 8 years; discounting at the rate of 5 per. 
cent. compound interest ? 

38. What. is the present worth of $1000, due in 20 
years; discounting at the rate of 6 per cent. compound 
interest? How much is the discount ? | 


XXXIV. 
ANNUITIES. 


An ANNUITY is a fixed sum of money payable periodi- 


cally, for a certain length of time, or during the life of | 


some person, or for ever. } 

Although the term annuity, in its proper sense, applies 
only to annual payments, yet payments which are made 
semiannually, quarterly, monthly, &c., are also called 
annuities. 

Pensions, salaries, and rents, come under the head of 
annuities. Annuities may, however, be purchased by the 
present payment of a sum of money. The party selling 
annuities, is usually an incorporated trust company, insti- 
tuted and regulated upon principles similar to those of an 
insurance company. ‘The company has an office, called 
an annutty office, where all its business is transacted. 

The present worth of an annuity which is to continue 
for ever, is that sum of money, which would yield an in- 
terest equal to the annuity. But the present worth of an 
annuity which is to terminate, is a sum, which, being put 
on compound interest, would, at the termination of the 
annuity, amount to just as much as the payments of the 
annuity would amount to, provided they should severally 
be put on compound interest, as they became due. . 

The sum to be paid for the purchase of a life annuity — 
which is the same as its present worth— depends not only 
upon the rate of interest, but, also upon the probable con- 
tinuance of the life or lives on which the annuity is grant- 
ed. In order to bring data of this kind into numbers, the 
bills of mortality in different places have been examined, 


ee 


| XXXIV. ANNUITIES. 197 


and from them, tables have been constructed, which show 
how many persons, upon an average, out of a certain 


number born, are left alive at the end of each year; and 
from these tables others have been constructed, showing 


the expected continuance of human life, at every age, 
according to probabilities. We shall not, however, treat 


the subject of life annuities in this work, and would refer 


: 


readers, who wish to become thoroughly acquainted with 
its theory, to the writings of. Simpson, De Moivre, Bai- 
ley, Price, and Milne. : 


PROBLEM I. ‘To find the amount of an annuity, which 
has been forborn for a given time. 
Before presenting the rule, let us inquire what would 


be the amount of an annuity of $100, forborn 4 years, 


allowing 5 per cent. compound interest? The last year’s 


payment will, obviously,.be $100 without interest; the 
last but one will be the amount of $100 for 1 year; the 


last but two will be the amount of $100 for 2 years; and 
so on: and the sum of the amounts will be the answer. 
Now the last payment with the amounts for the several 
years, form a series of continual proportionals. We, 


therefore, adopt the following— 


RULE. Find the sum of an increasing series of con- 


-tinual proportionals, whose first term is the annuity, whose 


ratio is the amount of 1 dollar for 1 year, and whose 
number of terms is the number of years. This sum is 
the amount. See Art. xxxit1, Problems 1st and 2nd. 

1. What is the amount of an annuity of $200, which 


has been forborn 14 years; allowing 6 per cent. interest ? 


2. What is the amount of an annuity of $50, which has 


been forborn 20 years; interest being 5 per cent.? 


3. What is the amount of an annual rent of $150, for 
born 7 years; allowing interest at 5 per cent.? 
4. If an annual rent of $1054 be in arrears 4 years - 


what is the amount, allowing 10 per cent. interest ? 


5. Suppose a person, who has a salary of $600 a year, 
payable quarterly, to allow it to remain unpaid for 3 years, 


how much would be due him; allowing quarterly com- 
pound interest at 6 per cent. per annum? 


Vie 


198 ARITHMETIC, XXXIV.” 


6. What is due on a pension of $150 a year, payable 
half-yearly, but forborn 2 years; allowing half-yearly com- 
pound interest, at 44 per cent. per annum? 

7. What is due on a pension of $300 a year, payable 
quarterly, but forborn 24 years; allowing quarterly com- 
pound interest, at.5 per cent. per annum. 


PROBLEM II. To find the present worth of an annuity 
which is to terminate in a given number of years. 

Before giving the rule, let us inquire, what is the pres- 
ent worth of an annuity of $100, to continue 4 years, 
allowing 5 per cent. interest? The present worth is, 
obviously, a sum, which, at compound interest, would 
produce an amount equal to the amount of the annuity. 
Now we can find the amount of any sum at compound 
interest, by multiplying the sum by the amount of 1 dollar 
for a year, as many times as there are years. Hence, to 
find a sum which will produce a given amount in a given 
time, we must reverse the process, and divide by the 


amount of 1 dollar for the time. Applying this principle — | 


to the example in question, we find by the preceding rule, 
that the amount of the annuity is $431. Then, dividing 
this amount by the amount of 1 dollar for 4 years, we find 
the present worth to be $354.593-+ 

RULE. Find the amount of the annuity as if it were in 


arrears for the whole time, and divide this amount by the — 


amount of 1 dollar at compound interest for the same 
time; the quotient will be the present worth. 

8. What is the present worth of an annuity of $500, 
to continue 10 years; interest being 6 per cent.? 

9. What is’ the present worth of an annuity of $80, to 
continue 22 years; interest being 5 per cent.? 


The operations in this rule being tedious, we introduce, 
upon the next page, a table, showing the present worth 
of $1 annuity, at 4, 5, 6, and 7 per cent., for every 
number of years, from 1 to 30. To find the present 
worth of an annuity by the use of this table, multiply the 


present worth of 1 dollar for the number of years, by the 
annuity. 


i 


) 


‘XXXIV. 


Y’rs.|4 per cent. 


7 per cent. 


— | SO Ss | Ea _———_ 


10 


12.1656 
12.6593 
13.1339 
13.5903 


14.0291 
14.4511 
14.8568 
15.2469 
15.6220 
"15.9827 
16.3295 
16.6630 
16.9837 
17.2920 


ANNUITIES. 
5 per cent. |6 per cent. 
.9523 9433 
1.8594 1.8333 
2.723 2.6730 
3.5459 3.4651 
4.3294 4.2123 
5.0756 4.9173 
5.7863 5.5823 
6.4632 6.2097 
fall © 6.8016 
T7217 7.3600 
8.3064 7.8868 
8.3632 8.3838 
9.3935 8.8526 
9.8986 9.2949 
10.3796 9.7122 
10.8377 | 10.1058 
11.2740 | 10.4772 
11.6595 | 10.8276 
L2:5883" |) hin set 
12.4622 | 11.4699 
12.8211 | 11.7640 
13.1630 | 12.0415 
13.4885 | 12.8033 
13.7986 | 12.5503 _ 
14.0939 | 12.7838 
14.3751 | 13.0031 
14.6430 | 13.2105 
14.8981 | 13.4061 
15.1410 | 18.5907 
15.3724 | 13.7648 


10. What is the present worth of an pari of $21.54, 
for 7 years; interest being 6 per cent.? 
11. What is the present worth of an annuity of 4 936, 
'e° 20 years, at 5 per cent.? 
12. What is the present worth of an annuity of $258. 
for 17 years, at 4 per cent.? 


i 


200 ARITHMETIC. XXXIV) 


13. Find the present worth.of an annuity of $'796.50,| 
to continue 28 years; interest being 7 per cent.? | 

14. A young man purchases a farm for $924; and) 
agrees to pay for it in the course of 7 years, paying 4 
part of the price at the end of eachyear. Allowing inter- 
est to be 6 per cent., how much cash in advance will 
pay the debt’. ee | 

15. Allowing interest to be 5 per cent., which will be} 
in my favor, to pay $15 a year for 10 years, or, to pay! 
$ 160 in advance ?—by how much? 


When an annuity does not commence until a given! 
time has elapsed, or some particular event has taken place, 
it is called a REVERSION. | 
_ PROBLEM Ht. ‘To find the present worth of an annuity) 
in reversion. 

RULE. Find, (by Problem 2nd.), the present value of | 
the annuity from the present time till the end of the period! 
of its continuance: find, also, its value for the time be- 
fore it is to commence: the difference of these two results’ 
will be the present worth. | | 

16. What is the present worth of an annuity of $200, | 
to be continued 7 years, but not to commence till 2 years | 
hence; interest being 6 per cent.? . ~ FI 

17. Find the present worth of a reversion of $152 a| 
year, to commence in 6 years, and to continue 18 years; | 
interest being 4 per cent. | 

As. What is the present worth of a reversion of $75 | 
a year, to commence in 5 years, and to continue 24) 
years; interest being 6 per cent.? - 

19. What must be paid for the purchase of a reversion 
of $450 a year, to commence in 5 years, and to continue | 
13 years; interest being 5 per cent.? | 

20. Find the present worth of a reversion of $942.30 | 
a year, to commence in 2 years, and to continue 11 | 
years; interest being 7 per cent. . | 

21. A father leaves to his son, a rent of $310 per’ 
annum, for 8 years, and, the reversion of the same rent — 
to his daughter for 14 years thereafter. What is the | 
present worth of the legacy of each, at 6 per cent.? © | 

We 


‘ : 


* 


i 
XXXV. ALLIGATION. 201 


| 22. What is the present worth of a reversion of $100 
ayear, to commence in 4 years, and to continue for ever; 
interest being 6 per cent.? 
_ This annuity continuing for ever, will, when it com- 
mences, be worth that sum of money which would yield 
$100 a year, at 6 per cent. interest. Therefore, after 
finding the principal, whose interest is $100 per annum, 
deduct from it a compound discount for 4 years; the re- 
mainder will be the present worth. / 
| 23. What is the present worth of a reversion of $824 
a year to commence in 7 eas and to continue for ever; 
mterest being 5 per cent.? 
_ 24. What is the present worth of a reversion of $530 
ayear, to commence in 22 years, and to continue for ever; 
interest 7 per cent.? 
_ 25. How much must be paid, at present, for a share 
ma fund, which, after the lapse of 20 years, will yield 
in income of $400 a year; interest 6 per cent.? 

26. How much must be paid, at present, for the title 
‘oan annuity of $1000, to commence in 40 years; inter- 
ast being 5 per cent.? ; 


XXXV. 
ALLIGATION. 


ALLIGATION relates to finding the mean value of a mix- 
ure composed of several ingredients of different values, 
ind is considered under two heads, viz. Alligation Medial, 
md Alligation Alternate. 


ALLIGATION MEDIAL. 


We rank under the head of Alligation medial, those 
juestions, in which the several ingredients and don re- 
‘pective values. are given, and the mean value of the 
-ompound is required. 

For example, a wine merchant bought several kinds of 
vine, as follows; 160 gallons at 40 cents per gallon; 74 


q 


| 


202 ARITHMETIC. XXXYj 


gallons at 60 cents per gallon; 225 gallons at 48 cenit 

per gallon; 40 gallons at 85 cents per gallon; and mixer| 

them together. It is required to find the cost of a gallor 

of the mixture. i | 

Now, if we find the whole cost of the several kind: 

of wine, and divide it by the whole number of gallons, i| 

is evident, that the quotient will be the cost of a single 

gallon of the mixture. | 

160 gallons, at 40 cents per gal., cost § 64.00 

75 gallons, at 60 cents per gal., cost $ 45.00 

225 gallons, at 48 cents per gal., cost $108.00 | 

40 gallons, at 85 cents per gal., cost $ 35.00 

500 the whole number of gallons, cost $252.00 

$ 252.00 + 500 =.504, or 5 cents and 4 mills. - 

Therefore, to find the mean value of a compound, 

composed of several ingredients, of different values, we 

give the following | 5 

RULE. Find the value of each ingredient, add these 

values together, and divide their sum by the sum of the 
ingredients. The quotient is the mean value. 


7 4 

1. A farmer mixed together 5 bushels of rye worth 70 
cents a bushel, and 10 bushels of corn worth 60 cents a 
bushel, and 5 bushels of wheat worth $1.10 a bushel, 
What is a bushel of the mixture worth ? ka 

2. A grocer mixed together 38]b. of tea at 50 cents a 
pound, 15; 1b. at 80 cents, 124]b. at 60 cents, 831b. at 
96 cents, 77]b. at 32 cents, and sold the mixture at a) 
profit of 20 per cent. At what price per pound did he. 
sell it ? . i 
3. A goldsmth melts together 11 ounces of gold 23. 
carats ane, 8 ounces 211 carats fine, 6 ounces of pure 
gold, and 2 ounces of alloy. How many carats fine is, 
the mixture? - 4 

We remark, that a carat is a 24th part. Thus, 23 
carats fine, means 33 of pure metal. Pure gold is 34. ; 
Alloy is considered of no value. >| 

4. On a certain day, the mercury in the thermometer . 
was Observed to stand, 2 hours at 60 degrees, 3 hours at | 


XXXV. ALLIGATION ALTERNATE. 203 
'§2°, 4 hours at 64°, 3 hours at 67°, l hour at 72°, and 
i" es at 75°. _ What was the mean temperature for that - 
ay: a 
5. A dealer bought 245 gallons of sirop at 34 cents a 
gallon, and 243 gallons at 38 cents a gallon, and mixed 
oth quantities “and 14 gallons of water together, and sold 
‘the mixture at a profit of 50 per cent. At what price per 
gallon did he sell it ? 
6nA goldsmith melts together 3 ounces of gold 18 
earats fine, 2 ounces 21 carats fine, and 1 ounce of pure 
la. What is the fineness of the compound ? 


ALLIGATION ALTERNATE. 


Under the head of Alligation Alternate are included 
those questions, in which the respective rates of the dif- 
ferent ingredients are given, to compose a mixture of a 
fixed rate. It is the reverse of Alligation Medial, and 
may be proved by it. . 

If we would find what quantities of two ingredients, 
different in value, would be required to make a com- 
pound of a fixed value, it is evident, that, when the value 
of the required compound exceeds that of one ingredient 
just as much as it falls short of the value of the other, 
we must take equal quantities of the ingredients to make 
the compound; because there is just as much lost on the 
one, as is gained on the other. 

If the value of the compound exceeds that of one in- 
gredient twice as much as it falls short of the value of the 
other, we must take ofthe ingredients in the ratio of $ to 
+, or 1 to 2. For instance, if we would mix wines, at 
4 dollars and 1 dollar a gallon, in such proportion that 
the mixture should be worth 2 dollars a gallon, we must 
take 1 gallon at 4 dollars to 2 gallons at 1 dollar; because 
there is just as much lost on 1 gallon at 4 dollars, as is 
gained on 2 gallons at-1 dollar. 

If we would mix wines, at 6 dollars and 2 dollars a 
gallon, in such proportion as would make the mixture 
worth 3 dollars a gallon, we should take of the two kinds 


| 


204 ARITHMETIC. XXXV_ 


{ 
in the ratio of } to 4, or 1 to 3; for, in this instance, 
there is as much lost on 1 gallon at 6 dollars, as is gainec 
on 3 gallons at 2 dollars. a | 

We see by the preceding ratios, that the nearer the 
value of the mixture is to that of one of the ingredients,| 
the greater must be the relative quantity of this ingredi-| 
ent, in forming the compound; and the farther the value 
of the mixture is from that of one of the ingredients, the’ 
less must be the -relative quantity of this ingredient in| 
making the compound. 4 

Hence, if we make the difference between the rate of 
each ingredient and that of the compound, the denomina- 
tor of a fraction having | for its numerator, these fracuons| 
express the ratio of the ingredients required to make the 
compound; and, when these fractions are reduced to ¢ 
common denominator, the numerators express the requir 
ed ratio of the ingredients. | 

If, for example, it be required to mix gold of 12 caratg| 
fine with gold of 22 carats fine, in such proportion thai| 
the mixture may be 18 carats fine, we can ascertain the| 
‘proportion of each kind in the following manner. The 
difference between 18 and 12 is 6; making 6 the de- 
nominator of a fraction with 1 for its numerator, we have 
the fraction 3; taking the difference between 18 and 22, 
we in like manner obtain the fraction 4; therefore, | 

| 


fractions, ¢ and 4, express the required proportion of 
each sort of gold. These fractions, when reduced to a 
common denominator, are 34 and 34, and the numerators| 
express the required proportion of each sort. Therefore, 
we must take 4 grains of 12 carats fine, and 6 grains of 
22 carats fine; or, in that ratio. | 

If, for a second example, we would make a mixture 
{8 carats fine from gold of 15 carats and 20 carats fine, 
we should, in the same manner, obtain the fractions, 2 
and 2, to express the required proportion of the two sorts 
of gold; consequently, in this instance, we should take 
2 grains of 15 carats fine, and 3 grains of 20 carats fine. 

Therefore, since the fineness of the compound is the 
same in both the preceding examples, if we would make 
a compound 18 carats fine, from the four kinds of gold 


XXXV. ALLIGATION ALTERNATE. 205 


mentioned in the two examples, we should take 4 grains 
of 12 carats, 6 grains of 22 carats, 2 grains of 15 carats, 
and 3 grains of 20 carats fine. : 

_ Now, these results may be readily obtamed by writing 
_ the rates of the given simples one under another, in reg- 
ular order, beginning either with the least or greatest, and 
alligating one of a less with one of a greater rate than that 
, of the compound, and writing the difference between 
| the rate of each simple and the rate of the compound, 
_ against the rate of the simple with which it is alligated. 


_ Thus, |12—— 4 grains 12 carats fine 
{ 15 2) CC 15 CC C6 
| Sag] 3 «20 
t ee a eA De, 


We may connect the rates of the simples differently, 
and obtain equally correct, but different results. 
Thus, |12—, 2 


7 


‘ 19/281 § 
23 3 


It must be observed, that the two simples linked to- 
_ gether, must always be one of a less, and the other of a 
greater rate, than the rate of the compound. : 
By connecting a less rate with a greater, and placing 
the differences between them and the mixture rate alter- 
nately, the gain on the one is precisely balanced by the 
loss on the other. This being true of every two, it is 
true of all the simples in the question, whatever may be 
their number. 
It is obvious, that a question in Alligation Medial admits : 
of a great variety of answers, all agreeing with the requi- 
_ sition of the question; for we may variously alligate the 
_ values of the ingredients, and thus obtain various results, 
all of which will be correct; and we may add all these 
_ together, and the results will be correct answers. We may 
_-also multiply, or divide the quantities found; for, if two 
quantities of two simples make a balance of loss and gain 
in relation to the value. of the compound, so must also the 
double or treble, the half or third part, or any other ratio 
of the quantities. 
18 


} 


{ 


206 ARITHMETIC. XXXV. : 


We shall give the questions in Alligation Alternate: 
under four cases. 4 


CASE I. The ratios of the several ingredients being 
given, to make a compound of a fixed rate. | 
RULE. First— Write the rates of the several ingredients | 
in a column under one another. i | 
2dly— Connect with a continued line the rate of each, 
ingredient less than the rate of the compound, with one or 
more rates greater than the rate of the compound; and 
each of a greater rate than the rate of the compound with. 
one or more of a less rate. | 
3dly— Write the difference between the rate of each ing | 
gredient and the rate of the compound, opposite the rate | 
of the ingredient with which it is connected. q 

Athly—If only one difference stand against any rate, 
ut will be the required quantity of the ingredient of that, 
rate; but, if there be several, their sum will be the quan-— 
tity required. | 

7. A goldsmith has gold of 17, 18, and 22 carats fine, , 
and also pure gold. What proportion of each sort must 
he take, to compose a mixture 21 carats fine? ) 

8. Having gold of 12, 16, 17, and 22 carats fine, what 
proportion of each kind must I take, to make a compound 
18 carats fine ? 

9. A merchant has spices at 30, 33, 67, and 86 cents 
a pound. How much of each sort must he take, to make 
& mixture worth 56 cents a pound ? 

10. A wine merchant has Canary wine at 50 cents a 
gallon, Sherry at 76 cents, and Claret at 175 cents per 
gallon. How much of each sort must he take, to make a 
mixture worth 87 cents a gallon? | 

11. A goldsmith wishes to mix gold of 16, 18, 19, 
and 23 carats fine, with pure gold, in such proportions 
that the composition may be 20 carats fine. What quan- 
tity of each must he take ? 

12. It is required to mix different sorts of wine, at 56, 
62, and 75 cents per gallon, with water, in such propor- 
tions that the mixture may be worth 60 cents a gallon. 
How much of each must be taken ? 


‘XXXV. ALLIGATION ALTERNATE. 207 


: 

| 13. How much corn at 52 cents a bushel, rye at 56 
eents, wheat at 90 cents, and wheat at 1 dollar a bushel, 
must be mixed together, that the composition may be 
‘worth 62 cents a bushel ? 

14. A silversmith wishes to mix alloy with silver of 
‘10, and 7 ounces fine, and pure silver, in such proportion 
‘that the mass may be 9 ounces fine: 12ozs. fine being 
pure. How much of each must he take? 


CASE II. When one of the ingredients is limited to a 
‘oertain quantity. 
RULE. Find the quantity of each ingredient, as in 
Case 1st. in the same manner, as though no such limitation 
‘were made; then as the difference against that simple, 
whose quantity is given, is to each of the other differences, 
$0 is the given quantity of that simple to the quantity re- 
“quired of each of the other simples. 

15. A trader has 90 pounds of tea worth 40 cents a 
: pound, which he would mix with some at 50 cents, some 
at 85 cents, and some at 90 cents. How much of each of 
the ciiier sorts must he mix with the 90 pounds, to make 
a mixture worth 60 cents a pound ? 


First solution. Second solution. 
40—— as 40— 25 
9|2°— 50—|-, 30 
60135 | 4 60 85— | 20 
9O0—— i | 90.4) 
mus 30 : 25—90:75 thus 25 : 3090 : 108 
a0 > 10— 90°: 30 25. % 20=3 90" --72S 
30 : 20=—90 : 60 25-1 0==90 + ..36 


30]b. at 85 cents, and 72\b. at 85 cents, and 
60 pounds at 90 cents. 36 pounds at 90 cents. 
16. A farmer wishes to mix corn at 54 cents a bushel, 
rye at 61 cents a bushel, and wheat at 96 cents a bushel, 
with 3 bushels of wheat worth I dollar and 10 cents a 
bushel. How much of each of the other three must be 
mixed with the 3 bushels of wheat at 1 dollar and 10 cents 
a bushel, that the mixture may be worth 75 cents a 


bushel ? 


Ans. 75 lb. at 50 cents, Sah 108 lb. at 50 cents, 


208 ARITHMETIC. | XXXV. 


17. How much gold of 16, 20, and 24 carats fine, and 
how much alloy, must be mixed with 10 ounces of 18 
carats fine, that the composition may be 22 carats fine ? 

18. How much silver of 6.5 ounces fine, and of 10.5 
ounces fine, and alloy, must be mixed with 17.1 ounces 
of pure silver, that the mass may be 9.5oz. fine? 

It must be observed, that pure silver is 12 ounces fine. 


CASE 111. When two or more of the ingredients are 
limited in quantity. . 


RULE. Find, as in Alligation Medial, what will be the | 
rate of a mixture made of the given quantities of the lim- | 


ited ingredients only; then consider this as the rate of a 
limited ingredient, whose quantity is the sum of the quan- 
tities of the limited ingredients, from which, and the rates 


of the unlimited ingredients, proceed to calculate the | 


several quantities required, as in Case 11. . 
~19. T have 18 gallons of wine at 48 cents a gallon, 8 
gallons at 52 cents, and 4 gallons at 85 cents, and would 
mix the whole with two other kinds of wine, one at a 
dollar and 26 cents, the other at 2 dollars and 12 cents 
a gallon. How much of the wine at a dollar and 26 cents, 
and of that at 2 dollars and 12 cents, must I mix with the 
other three, that the mixture may be worth a dollar a 

gallon ? 
18 gal. at .48- come to $8.64 ’ 


‘> ggal. at 62°“ 4.16 - 
4gal. at .85 3 3.40 
The 30 gal. come to 16.20, which is .54 a gallon. 


54 cents a gallon being the mean value of the 30 gallons, 
contained in the three kinds that are limited, I must now 
inquire how much of each of the other two sorts of wine 
at 1 dollar 26 cents, and 2 dollars 12 cents, must be 


mixed with 30 gallons at 54 cents a gallon, to make a 


mixture worth one dollar a gallon. 
| 54—, 26-4119 138 - 
POD iat. |-vi mie a aie 
Pohang 


gal. gal. gal. — gal. 


' Now as 188 : 46—30: 10 


) XXXV. ALLIGATION ALTERNATE. 209 


Therefore, I must take 10 gallons each of the two 
_ sorts, which are worth 1 dollar 26 cents, and 2 dollars 12 
cents a gallon. 
20. How much gold of 14 and 16 carats fine must be 
_ mixed with 6 ounces of 19, and 120z. of 22 carats fine, 
that the composition may be 20 carats fine ° 
21. A silversmith has silver of 6, 7, and 9 ounces fine, 
_ which he wishes to mix with 9 ounces of 10 ounces fine, 
and 9 ounces of pure silver, to make a mass, that shall 
be 8 ounces fine. How much of each of the three first 
must he take? 
22. A lady purchases 7 yards of eelcor at 22 cents a 
_ yard, and 7 yards at 20 cents a yard, and wishes to know 
_ how many yards of two other kinds, one at 16 cents and 
the other at 17 cents a yard, she must purchase, to make 
the average price of the whole 18 cents a yard. Find 
the two quantities. 


CASE IV. When the whole compound is limited to a 
certain quantity. 

RULE. Find an answer, as in Case 1, by alligating 
then, as the sum of the quantities thus found, is to the 
given quantity, so is the quantity of each ingredient found 
by alligating, to the required quantity of it. 

23. A goldsmith has gold of 15, 17, 20, and 22 carats 
fine; and would melt together of all these sorts so much, 
as to make a mass of 40 ounces 18 carats fine. How much 
of each sort is required ? 


15—— E Or thus we. 

jd ee 

“aca - sae 3 

22—— j A) Torres phon ; 
10:40 == 4 216 10:40=—2: 8 
10:40—2: 8 10 : 40=—4: 16 
10:3.40=1;.4 10.: 40==3.2.12 
10 : 40=3: 12. 10:40—1: 4 


Ans. 1602. of 15; 802. of 17; 402. of 20; and 1202. 
| of 22 carats fine. 


24. Having three sorts of raisins at 9, 12, and 18 cents 
is* 


ILO ARITHMETIC. XXXVI 


a pound, what quantity of each sort must I take, to fill 
a cask containing 210 pounds, that its contents may be | 


worth 14 cents a pound? 


25. Of four different kinds of apples at 31, 37, 46, and | 
74 cents a bushel, what quantity of each must be taken, 


to fill a bin containing 9 bushels, to make its contents 
worth 50 cents a bushel? 


XXXVI. 
PERMUTATIONS. 


PERMUTATION—which is also called variation—means 
the different ways in which the order or relative position 
of any given number of things may be changed. The 
only object to be regarded in Permutation, is the order in 
which the things are placed; for no two arrangements are 
to have all the quantities in the same relative position. 

For example, two things, a, and b, are capable of only 
two changes in their relative position, viz. ab, ba; and 
this number of changes is expressed by 1X2; ‘but three 
things, a, b, and c, are capable of six variation: viz. 
abc,acb, bac, bca,cab, cba, and this number 
of permutations is expressed by 123; and four things, 


a, b, c, and d, are capable of 24 variations, viz, abc d, - 


abdc,acbd;acdb, adbct, adc b; 6b aca, bes 
de, bead, becda, bdac, bdca; cabd;,cadp, 
c ba die b'da) cid. ab; c:d boas da 0 cy 2 eee 
ac, dbca, dcab, sidan: and this number of per- 
mutations 1s expressed by A Mehl ote 

In like manner, when there are 5 things, every four of 
them, leaving out the 5th, will have 24 variations; con- 


———— 


sequently by taking in the 5th, there will Ve 5 times 24 _ 


variations. 


PROBLEM I. ‘To find the number of permutations that 
can be made of any given number of things, all different 


from each other. 
RULE. Multiply the terms of the A Ee series of 


; 
| 


4 

] 

} 

! va 
; 

5 . 


| XXXVI. PERMUTATIONS. | 211 


numbers, from 1 up to the given number of things, con- 
-_tinually together, and the product will be the answer. 


1. How many changes can be made in the order of 
the six letters, abc def? 

2. How many changes may be rung on seven bells ? 

3. Five gentlemen agreed to board together, as long 
as they could seat themselves every day in a different 
position at the dinner table. How long did they board 
together ? 

4. How many changes may be made in ‘the order of 
the words in the following verse? Proci tot tibi sunt, 
virgo, quot sidera coelo. 

_ 5. How many different sums of dollars can be expres- 
sed by the nine digits, without using any one of them more 
than once in the same sum ? 

6. How many different arrangements may be made in 
seating a class of 20 scholars ? 

7. A gentleman, who had a wife and eight daughters, 
one day said to his wife, that he intended to arrange the 
family in a different order every day at the dinner table, 
and that he would never give one of his daughters in 
marriage, till he had completed all the different arrange- 
ments of which the family was capable. How many years 
from that day must elapse, before either of his daughters 
can be married ? 


_ When several of the things are of one sort, and several 
of another, &c. the changes that can be made upon the 
whole is not so great, as when all the things are different. 
For instance, we have seen that the letters a b c admit 
of six variations; but, if two of the quantities be alike, as 
aa b,-the six variations are “~duced to three, aa b, baa, 


aba, which may be expressed by =X". We have also 


seen that the letters ab c d admit of 24 variations; but 
if we have aa b b, the 24 variations are reduced to six, 
im. aabwv,abba,abab, bbaa, baab, bab a, 
‘and this number of variations may be expressed by 


(1xX2x3xK4 : Sele, ae. 
Beoxciscr. “Hence, we have, as follows,— | 


| 
* | 


212 ARITHMETIC. XXXVI 


PROBLEM I. To find the number of changes that may | 
be made in the arrangement of a given number of things, 
whereof there are several things of one sort, several of 
another, &c. | 

RULE. Take the natural series of numbers from 1 up 
to the given number of things, as if they were all differ- 
ent, and find the product of the terms. 

Then take the natural series from 1 up to the number 
of similar things of one sort, and the same series up to” 
the number of similar things of a second sort, §c., and 
divide the first product by the joint product of all these 
series, and the quotient will be the answer. 

8. Find how many changes can be made in the order 
of the letters a aa bbc. 

If the letters in this question were all different, they 
would admit of 1X2X3xX4x5xX6 =720 variations; but 
since a is found 3 times, we must divide that number of 
variations by | <2X3; and, since b occurs twice, we must 
again aivide by 1X2; therefore the number of variations 
will be ee = 60. ; 

9. How: many changes can be made in the order of 
the lettersaaabbbbecdee? 

10. How many variations may take place in the suc- 
cession of the following musical notes, fa, fa, fa, sol, sol, 
la, mi, fa? 

11. How many whole numbers can you make out of — 
the number 1220055055, using all the figures each time? 

12. How many variations can be made in the order 
of the figures in the number 97298279289 ? 


PROBLEM IH. Any number of different things being 
given, to find how many changes can be made out of them, 
by taking a given number of the things at a time. 

RULE. Take a series of numbers commencing with the 
given number of things and decreasing by 1, till the 
number of terms is equal to the number of things to be 
taken at a time, and the product of all the terms of this’ 
series will be the answer. | 

To illustrate the rule, we will take the four letters 
abed, and find the number of variations that can be 


“XXXVIL COMBINATIONS. 213 


made upon them, by taking two at atime. In the first 
place, we will write the letter a on the left hand of each 
of the other letters, and the variations will be three, viz. 
ab, ac, ad; we will do the same with each of the other 
letters, thus, b a, bc, bd; ca,cb,cd; da,db,dc. 
Now we have all the changes that can be made upon the 
four letters, taking two at a time, and they are 4X3=12. 
__ We will also find, in the same manner, how many 
changes can be made on the same four letters, by taking 
three at a time; writing @ on the left, thus, a b c, a b d; 
achb,acd; adb,adc, we have 3X2=6 variations. 
Now, since each of the letters is to be written in the same 
manner on the left, we shall have four such classes of 
variations, and the whole number will be 4x*3X*2—=24 
variations. 

__ 13. How many changes can be made upon the letters 
abcdef, by taking three at a time? 

14. How many different whole numbers can-be ex- 
pressed by the nine digits, by using two at a time? 
15. How many different whole numbers can be ex- 
pressed by the nine digits, by using four at a time? 
16. How many different numbers can you express 
‘with the nine digits and a cipher, by using five at a time? 


: 
XXXVII. 
COMBINATIONS. 


CoMBINATION consists in taking a less number of things 
out of a greater without any regard to the order in which 
they stand. This is sometimes called Election or Choice. 

No two combinations can have the same quantities; for 
imstance, the quantities, a and b, admit of only one com- 
bination, because a b and b a are composed of the same 
quantities; but, if a third quantity ¢ be added, we can 
imake three combinations of two quantities out of them, 
‘because the third quantity e may be added to each of the 
two former, thus, a b, a c, b c; this number of combina- 


: 3 
tions may be expressed by 


— If we adda fourth letter, 


Ae i 
| 
{ 


214 ARITHMETIC. XXXVI 


d, we can make six combinations of two letters out of the | 
four, since the new quantity d, may be combined with. 
each of the former ones; thus, ab, ac, bc, ad, bd, 
cd; and this number of combinations may be expressed | 


4x3 
by Tyas : 

If we would make a combination of four, it is evident. 
that only one such combination can be made out of the. 
letters a bc d; but if a fifth letter, e, be added, we can | 
make five such combinations; thus, abcd, abce,' 


abed,aecd,bede; and this number of combina-— 


5X 4X 3X2 
tions may be expressed by 337: 


PROBLEM 1. ‘To find the number of combinations from | 
any given number of things, all different from each other, 
taking a given number at a time. | 

RULE. Take a series of numbers, the first term of which | 
is equal to the number of things out of which the combi-_ 
nations are to be made, and decreasing by 1, till the num~ 
ber of terms is equal to the number of things to be taken | 
at a time, and find the product of all the terms. 

Then take the natural series 1, 2, 3, &c. up to the | 
number of things to be taken at a time, and find the pro 
duct of all the terms of this series. 4 

Divide the former product by the latter, and the quo- 
trent will be the answer. a 


1. How many combinations of 3 letters can be made 
out of the 6 letters abc def? 4 

2. How many different yoke of oxen may be selected _ 
from twelve oxen? . 

3. How many different span of horses can be selected 
from eighteen horses ? é | 

4. A drover agreed with a farmer for a dozen sheep, — 
to be selected out of a flock of two dozen; but while he; 
was making the selection, the farmer told him, he might, 
take the whole flock, if he would give him a cent for. 
every different dozen that could be selected from it. To | 
this the drover readily agreed. How many dollars did | 
the whole flock come to, at that rate ? 

5. A general, who had often been successful in war, 


Se — er ea 


XXXVII. COMBINATIONS. 215 


was asked by his king, what reward he should confer upon 
him for his services. The general only desired a farthing 
for every file, of 10 men in a file, which he could make 
with a body of 100 men. How much did the general’s 
modest request amount to ? 


PROBLEM H. ‘To find the various combinations of a 
given number of things, which may be made out of an 
equal number of sets of different things, one from each 
set. 

RULE. Multiply the number of things in the several 
sets continually together, and the product will be the 
answer. ; 

A combination of this kind is called the composition 
of quantities. The rule may be illustrated thus. If there 
are only two sets, and we combine every quantity of one 
set with every quantity of the other set, we shall make 
all the compositions of two things in these two sets; and 
the number of compositions is evidently the product of 
the number of things in one’ set by the number of things 
in the other set. Again, if there are three sets, then the 
compositions of two in any two of the sets, being com- 
bined with every quantity of the third set, will make all 
the compositions of three in the three sets. That is, the 
compositions of two in any two of the sets, being multi- 
plied by the number of things in the third set, will give 
all the compositions of three in the three sets; and this 
1esult is the joint product of all the numbers in the three 
sets. , 

6. Suppose there are four companies, in each of which 
there are 9 men; in how many ways can 4 men be chosen, 
one out of each company ? 

7. Suppose there are five parties, at one of which 
there are 6 young ladies, at another 8, at a third 5, at a 
fourth 7, at a fifth 10. How many choices are there, in 
selecting 5 young ladies, one from each party? 

_ 8. How many changes are there in throwing fou 
dice, each die having six sides ? : 

9. A certain farmer has 5 barns, in one of which he 
has 15 cows, in another 11, in another 5, in another 9, 


216 ARITHMETIC. XXXVI. 


and in another 7. How many different selections may 


be made, in choosing 5 cows, one from each barn ? 


10. How many variations can be made in selecting a 
flock of a dozen sheep from 12 folds, one from every) 


fold, in each of which there are 10 sheep? 

11. In a certain school there are seven classes, the 
first containing 12 boys, the second 7, the third 9, the 
fourth 10, the fifth 11, the sixth 8, and the seventh 13. 
How many variations can be made in selecting 7 boys, 
one from each class ? 


XXXVIII. 
EXCHANGE. 


| 
j 
' 
: 
| 


| 
| 


| 
| 
| 
| 
| 


Scholars, who are to prosecute a course of classical studies, and those, 


who are not expected to engage in any extensive mercantile business, may 
omit the exercises in this article. 


ExcuancE is the act of paying or receiving the money 
of one country for its equivalent in the money of another 


country, by means of Bills of Exchange. 'This operation, 


therefore, comprehends both the reduction of moneys 


and the negotiation of bills; it determines the comparative 
value of the currencies of different nations, and shows how 
foreign debts are discharged, and remittances made from 
one country to another, without the risk, trouble, or 
expense of transporting specie or bullion. 


A Bill of Exchange is a written order for the payment 
of a certain sum of money, at an appointed time. It is 


| 


a mercantile contract, in which four persons are mostly 


concerned, as follows. 
First—The Drawer, who receives the value, and is.also 
called the maker and seller of the Bill. 
Second—The debtor in a distant place, upon whom the 
Bill is drawn, and who is called the Drawee. He also is 


called the Acceptor, after he accepts the Bil, which is an 


engagement to pay it when due. 
Third—The person who gives the value for the Bill, 
who is called the Buyer, Taker and Remitter. 


| 


XXXVILII. EXCHANGE. Q17 


I Fourth—The person to whom the bill is ordered to be 
| paid, who is called the Payee, and who may, by endorse- 
| ment, pass it to any other person. 

' Many mercantile payments are made in Bills of Ex- 
| change, which pass from hand to hand, until due, like 
any other circulating medium; and the person who at any 
) time has a Bill in his possession, is called the holder. 

_ ~-To transfer a Bill payable to order, the payee should 
express his order of paying to another person, which 1s 
always done hy an endorsement on the back of the Bill. 

_ An endorsement may be blank. or special. A blank 
endorsement consists only of the endorser’s name, and 
the Bill then becomes transferable by simple delivery. 
A special endorsement orders the money to be paid to a 
particular person, who is called the endorsee, who must 
also endorse the Bill, if he negotiates it. A blank en- 
-dorsement may always be filled up with any person’s 
fame, so as to make it special. Any person may endorse 
-a Bill, and every endorser, as well as the acceptor, is a 
“security for the Bul, aud may be sucd for payment. 

In reckoning when a Bill, payable after date, becomes 
“due, the day on which it is dated, is not included. When 
the time is expressed in months, calendar months are un- 
derstood; and when a month is longer than the succeed- 
ag it js a rule not to go, m the computation, into a third 
‘month. Thus, if a Bill be dated the 28th, 29th, 30th, 
‘or 31st, of January, and payable one month after date, 


the term equally expires on the last day of February. 
' An endorsement may take place at any time after the 
Bill is issued, even after the day of payment is elapsed. 
' When the holder of a Bill dies, his executors may en- 
dorse it; but, by so doing, they become answerable to 
their endorsee personally, and not as executors. 
A Bill payable to bearer is transferred by simple deliv- 
ery, without any endorsement. 
_ Bills should be presented for acceptance, as well as 
| for payment, during the usual hours of business. 
The common way of accepting a Bill is for the drawee 
to write his name at the bottom or across the body of 
it, with the word, accepted. 
19 


| 


218 ARITHMETIC. XXX VII 


When acceptance or payment has been refused, the 
Holder of the Bill should give regular and immediate 
notice to all the parties, to whom he intends to resort for 
payment; for if he do not, they will not be liable to pay. | 

With respect to the manner, in which notices of non- 
acceptance or non-payment are to be given, a difference. 
exists between Inland and Foreign Bills. | 

In the case of Foreign Bills, a Protest is indispensa-_ 
bly necessary: thus, a Public Notary appears with the 
Bill, and demands either acceptance or payment (as the 
case may be;) and on being refused, he draws up an in- 
strument, called a Protest, expressing that acceptance or 
payment (as the case may be) has been demanded and. 
refused, and that the holder of the Billintends to recover 
any damages which he may sustain in consequence. This 
instrument is admitted, in foreign countries, as a legal 
_ proof of the fact. 

The Protest on a Foreign Bill should be sent as soon 
as possible, to the drawer or negotiator; and if it be for 
non-payment, the Bill must be sent with the Protest. 

A Protest is not absolutely necessary to entitle the 
holder to recover the amount of an Inland Bill from the 
drawer or endorser: it is sufficient if he give notice, by 
letter or otherwise, that acceptance or payment (as the 
case may be) has been refused, and that he does not 
mean to give credit to the drawee. 

If the person, who is to accept, has absconded, or 
cannot be found at the place mentioned in the Bill, Pro- 
test is to be made, and notice given, in the same manner 
as if acceptance had been refused. 

It is customary, as a precaution against accident or 
miscarriage, to draw three copies of a Foreign Bill, and to 
send them by different conveyances. They are denomi- 
nated the First, Second, and Third of Exchange; and 
when any one of them is paid, the rest become void. 

When acceptance is refused, and the Bill is returned 
by Protest, an action may be commenced immediately 
against the Drawer, though the regular time of payment 
be not arrived. His debt, in such case, is considered 
as contracted the moment the Bill is drawn. 


XXXVIII. EX CHANGE. 219 
FORM OF A BILL OF EXCHANGE. 


New York, Nov. 4, 1834. 
Excuance for £3000 sterling. 

At thirty days sight of this, my first of Exchange, 
(second and third of the same tenor and date not paid), 
pay to Robert N. Foster, or order, Three Thousand 
Pounds Sterling, with or without further advice from me. 

' Enwin D. Harper. 
Messrs. Knox and Farnuam, 
Merchants, London. 


InLanD Excuance relates only to remitting Bills from 
one commercial place to another in the same country; by 
which means debts are discharged more conveniently than 
by cash remittances. 

Suppose, for example, A of New Orleans is creditor 
to B of Boston 1000 dollars, and C of New Orleans is 
debtor to D of Boston 1000 dollars; both these debts 

-may be discharged by means of one Bill. Thus, A draws 
for this sum on B, and sells his Bill to ©, who remits it 
to D, and the latter receives the amount, when due, from 
‘B. Thus, by a transfer of claims, the New Orleans 
debtor pays the New Orleans creditor, and the Boston 
debtor the Boston creditor, and no money is sent from 
“one place to the other. This business is usually con- 
ducted through the medium of Banks, which are in the 
habit of buying both foreign and inland Bills of exchange, 
and transmitting them to the places on which they are 
drawn for acceptance. 

Inland Bills of Exchange are sometimes called Drafts; 
and the following short form of the instrument is adopted. 


: 
: 


| 
: 
: $ 56070, Boston, Dec. 8, 1834. 
Three months after date, pay to the order of 
Charles S. Hooper, Five hundred dollars 50 cents, value 
received, and charge the same to account of 

' Hannum & Lorine. 
To SteruEN FroTHINGHAM, 


Merchant, Norfolk. 


220 ARITHMETIC. XXX VITL | 


Some explanation of mercantile language used in rela-_ 
tion to Bills of Exchange seems necessary, that the | 
learner may have a clear idea of the questions, which will 
be given for practice. 

When a merchant in the United States draws on his | 
banker in London, his draft is styled ‘‘ Bill on London” 
or ‘‘ United States on London;’’ and if he sells his Bill 
at more than adollar for 54d. sterling, the exchange is — 
said to be above par; and if ‘he sells at less than a dollar | 
for 54d. sterling, below par: if a merchant in London 
draws on his banker in the United States, his draft is 
styled ‘‘ London on United States;’’ and if he sells his— 
Bilkat more than 54 d. sterling for the dollar, the exchange 
is said to be above par; and if he sells at less than 54d. 
sterling for the dollar, below par. / . 

If the merchant in London draws on his banker in Paris, 
it is ** London on Paris,’’ or ‘‘ London on France.” If 
the merchant in Charleston, S. C. draws on his banker in 
New York, it is ‘‘ Charleston on New York.”’ &c. | 


GREAT BRITAIN. 


{n Great Britain, accounts are kept in pounds, shillings, 
pence, and farthings, Sterling. 

The par value of the United States dollar is 4s. 6d. 
sterling; therefore, the dollar is equal to 7 of a pound 
sterling. Hence, any sum of sterling money, (the shil- 
lings and pence, if any, being expressed in a decimal,) 
may be reduced to Federal money, by multiplymg by 40 — 
and dividmg by 9: and, any sum in Federal money may 
be reduced to sterling money, by multiplying by 9 and 
dividing by 40. Or, if sterling money be increased by 
4 of itself, the sum expresses the same value in the old 
currency of New England: and, conversely, if the old 
currency of New England, be decreased by i of itself, 
the result is the expression in sterling money. @ 

1. United States on London. Reduce £784 14s. _ 
103d. sterling to Federal money, at par. 

2. London on United States. Reduce 3487 dollars 
75 cents to sterling, at par. : 


| XXXVIIL. EXCHANGE. 221 


8. United States on London. Reduce £2006 11s. 


 sterlmg to Federal money; exchange at 4 per cent. below 


par. 
4. London on United States. Reduce 4287 dollars 
50 cents to sterling; exchange at 4 per cent. above par. 
5. London on United States. Reduce 3646 dollars 


50 cents to sterling; exchange at 2 per cent. below par. 


6. United States on London: Reduce £4109 Ils. 
10d. sterling to United States currency; exchange at 7 
per cent. above par. 

7. United States on London. Reduce £5129 15s. 


6d. sterling to Federal money; exchange at 5 per cent. 
above par. 


The law assimilating the currency of Ireland to that af 


England, took effect in January 1826. ~ All invoices, 
contracts, &c. are considered there, in law, British cur- 


rency, unless otherwise expressed. 

8. United States on Dublin (Ireland). Reduce £1834 
2s. 104d. sterling to Federal money; exchange at 4 per 
cent. above par. 


FRANCE. 


Accounts were kept in France previous to 1795, ac- 

cording to the old system, in livres, sous, and deniers. 
12 deniers= 1 sol or sou; 
20 sous =1 livre; 
6 livres ‘ = 1 ecuor crown, silver. 

By the new system, accounts are kept in francs, 

decimes, and centimes. 
10 centimes = 1 decime. 
10 decimes = 1 franc. 
The value of the franc is 182 cents in Federal money. 
80 francs ==81 livres. 

9. United Stateson France. Reduce 7232 francs 38 
centimes to Federal money; exchange at 1 dollar for 5 
francs 30 centimes. 

10. France on United States. .Reduce 4093 dollars 
80 cents to money of France; exchange at 5 francs 30 


~ centimes for the dollar. 


ioe 


pd). ARITHMETIC. XXXVIIL.- 


11. France on United States. Reduce 1834 dollars 


65 cents to French currency; exchange at 5 francs 40— 


centimes for a dollar. 
12. United States on France. Reduce 20828 frances 


67 centimes to Federal money; exchange at 1 dollar for — 


5 francs 38 centimes. 


13. United States on France. Reduce 12893 francs © 


27 centimes to Federal money; exchange at 1 dollar for 
5 francs 33 centimes. 


HAMBURGH. 


Accounts are kept here in marks, scnillings or sows, 
and pfenirgs, Lubs. 
12 pfenirgs = 1 sol or schilling, Lubs; 
16 schillings—=1 mark, Lubs; 
3 marks 1 reichsthaler or rix dollar specie. 


Accounts are also kept, particularly in exchanges, in 


pounds, shillings, and pence, Flemish. 
12 pence or grotes= 1 shilling. 
20 shillings =1 pound, Flemish. 
The word Lubs originally meant money of Lubeck, 
which is the same with that of Hamburgh, and the term 
is intended to distinguish this money from the Flemish 


denominations, and also from the money of Denmark and ~ 


other neighboring places. 

The mark Lubs is worth 22 shillings Flemish, or 32 
grotes; consequently the sol Lubs is 2 grotes Flemish, 
and the shilling Flemish 6 schillings Lubs. 

Banco or bank money, in which exchanges are reckon- 
ed, and curréncy, are the two principal kinds of money. 

Banco consists of the sums of money deposited by 


merchants and others i the bank, and inscribed in its — 


books; which sums are not commonly drawn out, but are 


transferred from one person to another in payment of a 


debt or contract. 


Current money, or currency, consists of the common ~ 


coins of the city, in which expenses are mostly paid. 


_ The bank money is more valuable than currency, and ~ 
bears a premium varying from 18 to 25 per cent. This ~ 


XX XVIII. EX CHANGE. 223 


premium is called the agio. For instance, when the agio 
is 20 per cent. 100 marks banco are valued at 120 marks 
currency. 

The mark banco is valued in the United States at 334 
cents. 

14. United StatesonHamburgh. Reduce 1148 marks, 
5 schillings, 4 pfenirgs banco to Federal money; exchange 
at 33 cents per mark banco. 

15. Hamburgh on United States. Reduce 1245 dol- 
lars 75 cents to money of Hamburgh; exchange at 3 
marks banco per dollar. 

16. United StatesonHamburgh. Reduce 6194 marks 
12schillings banco to Federal money; exchange at 34 cents 
per mark banco. 

17. United Stateson Hamburgh. Reduce8246 marks 
8 schillings banco to Federal money; exchange at 35 cents 
per mark banco. 

18. Hamburgh on United States. Reduce 757 dollars 
90 cents to money of Hamburgh; exchange at 1 mark 
banco for 33 cents. 


AMSTERDAM AND ANTWERP. 


In these places accounts were formerly kept in florins, 
Stivers, and penning; or in pounds, shillings, and pence, 
Flemish. 

16 pennings= 1 stiver, 


20 stivers ==1 florin or guilder. 

In Flemish, 12 grotes or pence, or 6 stivers— 1 shilling, 
20 shillings, or 6 florins ==I pound; — 
24 florins, or 50 stivers = I rix dollar. 


By the new system, adopted in 1815, accounts are 
kept throughout the kingdom of the Netherlands in florins 
or guilders, and cents. . 

. 100 cents 1 florin or guilder. 

The par value of the florin, in Federal currency, is 40 
cents. 

19. United States on Amsterdam. Reduce 13790 
‘florins 15 stivers to Federal money; exchange at 36 cents 
per florin. 


224 ; ARITHMETIC. XXX VILL. | 


20. United States on Antwerp. Reduce 6281 florins 
88 cents to Federal money; exchange at 40 cents per 
florin. | 

21. Amsterdam on United States. Reduce 2482 dol-_ 
lars 334 cents to Dutch money; exchange at 36 cents per 
florin. 

22. Antwerp on United States. Reduce 3436 dollars 
72 cents to Dutch money; exchange at 38 cents per 
florin. . 

23. United States on Antwerp. Reduce 7294 florins 
50 cents to Federal money; exchange at 42 cents per 
florin. : 

24. United States on Amsterdam. Reduce 10148 
florins to Federal money; exchange at 41 cents per florin. 


PORTUGAL. 


In Portugal, accounts are kept in milrees and rees; and 
also in old crusados. | 
1000 rees—=1 milree. . 
400 rees=1 old crusado or crusado of exchange. 
480 rees—=1 new crusado. 

There are three sorts of money used in Portugal; viz. 
effective money, i. e€. specie; paper money, which is ata 
discount; and legal money, consisting of half specie and 
half paper. 

The value of the milree in Federal money is 1 dollar 
24 cents. d 

25. United States on Lisbon. Reduce 964 milrees 
475 rees to Federal money; exchange at 1 dollar 24 cents 
per milree. 

26. Lisbon on United States. Reduce 1274 dollars 
66 cents to money of Portugal; exchange at 1 dollar 25 
cents per milree. | 

27. United States on Portugal. Reduce 1248 milrees” 
645rees to Federal money; exchange at 1 dollar 26 cents 
per milree. a 

28. United States on Portugal. Reduce 1846 milrees 
500 rees to Federal money; exchange at 1 dollar 23 cents 
per milree. 


XXXVIII. EXCHANGE. 225 


The exchanges of Brazil, in South America, are similar 
to those of Portugal; there is, however, a difference in 
the value of their moneys; that of Portugal is half specie 
and half paper, called legal money, and that of Brazil is 
effective. 


SPAIN. 


The most general mode of keeping accounts in Spain 
is in reals, of 34 maravedis; but there are nine different 
reals, each divided into 34 maravedis, but differing in 
value. Four of these reals are of general application, 
and five of local use. . 

The four principal moneys of Spain are the real vellon, 
the real of old plate, the real of new plate, and the real 
of Mexican plate; and in order to obtain a distinct view 
of them, it may be proper to make the real vellon the 
basis of all the rest. 

The real vellon is the twentieth part of the hard dollar, 
(peso duro), universally known by the name of the Span- 
ish dollar, which is the same in value with the dollar of 
the United States. 

The division of the real vellon is into quartos, ochavos, 
and maravedis. 

2 maravedis—1 ochavo, 

2 ochavos =1 quarto, 
8 quartos, or 34 maravedis=1 real vellon; 
but maravedis are commonly used to express any fraction 
ofa real: thus, we say 1 real 33 maravedis. 
_ The real of old plate is better than the real vellon, in 
the proportion of 32 to 17. Thus 17 maravedis of old 
plate are equal to 32 maravedis vellon; the quartos and 
ochavos are in the same preportion. The real of old 
plate is not a coin; it is the most general money of ex- 
change. 8 of these reals make the piastre, which is also 
ealled the dollar of exchange. 102 of these reals are 
equal to the hard dollar. When plate only is mentioned, 
old plate is understood. 
'- The real of new plate is double the real vellon; there- 
fore 34 maravedis of new plate are equal to 68 of vellon; 


) 
| 
| 
| 


| 
| 
| 
226 ARITHMETIC. XXXVI 


quartos and ochavos in proportion. This real is a coin, 
but not a money of account in any general way; it is the 
tenth part of the hard dollar, and is estimated in the 
United States at 10 cents, and the real vellon at 5 cents. 

The real of Mexican plate is divided into halves and 
quarters, called medio and quartillo. It is the eighth part 
of the hard dollar, and is the chief money of account in 
Spanish America, where it is divided into sixteenths. 

The doubloon de plata, or doubloon of exchange is four 
times the value of the piastre, or dollar of exchange. 

The ducado de plata, or ducat of exchange is worth 
11 reals 1 maravedi old plate, or 20 reals 2513 maravedis 
vellon. 

29. United States on Spain. Reduce 3148 dollars 
(of exchange) 6 reals 32 maravedis plate. to Federal 
money, exchange at 67 cents per piastre ? 

30. Spain on United States. Reduce 1821. dollars 
60 cents to Spanish money; exchange at 68 cents per 
dollar of exchange. 

31. United States on Spain. Reduce 1286 doHars 
(of exchange) 7 reals 17 maravedis plete to Federal money; 
exchange at 64 cents per dollar of exchange. 

32. United States on Spain. Reduce 2136 doubloons 
of exchange to Federal money; exchange at 68 cents per 
dollar (of exchange). 

33. United States on Buenos Ayres. Reduce 4680 
rials of Mexican plate to Federal money; exchange at 12 
cents per rial. | 


SWEDEN. 


In Sweden, accounts are kept in rix dollars specie, 
skillings, and rundstycken or ore. 
12 rundstycken or ore=1 skilling; 
48 skillings —=1 rix dollar specie. : 
The Swedish dollar agrees in value with the dollar of 
the United States. . 
34. United States on Sweden. Reduce 3955 rix dol- 
Jars 24 skillings to Federal money; exchange at 1 dollar 
2 cents fer rix dollar. . 


XXX VIII. EXCHANGE, 227 


35. Sweden on the United States. Reduce 1344 
dollars 87 cents Federal money to Swedish money; ex- 
change at 1 rix dollar for 1 dollar 2 cents. 

36. United States on Sweden. Reduce 2481 rix 
dollars 36 skillings to Federal money; exchange at 1 dollar 
per rix dollar. 

37. Sweden on the United States. Reduce 819 dol- 
lars 873 cents Federal money to Swedish currency; ex- 
change at 1 rix dollar per dollar. 

38. United States on Sweden. Reduce 1234 rix 
dollars 12} skillings to Federal money; exchange at 98 
cents per rix dollar. 

39. United States on Sweden. Reduce 1126 rix 

‘dollars 42 skillings to Federal money; exchange at 1 dol- 
lar 3 cents per rix dollar. 


RUSSIA. 


In Russia accounts are kept in rouoles and copecks. 

The rouble is also divided into 10 grieven. 
10 copecks==1 grieve or grievener, 
: 10 grieven or 100 copecks, = 1 rouble. 

The silver rouble is estimated in the United States at 
75 cents; but the commercial business of Russia is car- 
ied on, in a paper currency much inferior to that of 
specie. The variable agio of the paper, substituted for 
the silver rouble, makes exchanges with Russia extreme- 
ly fluctuating, as the paper rouble improves or declines 
in value. 

40. United States on Russia. Reduce 4182 roubles 
64 copecks to Federal money; exchange at 25 cents per 
rouble. 

41. Russia on the United States. Reduce 2614 dol- 
lars 15 cents to Russian money; exchange at 1 rouble for 
25 cents. 

42. United States on Russia. Reduce 5416 roubles 
50 copecks to Federal money; exchange at 28 cents per 
‘rouble. 

43. Russia on United States. Reduce $3148.56 to 
Russian currency; exchange at 1 rouble for 30 cents. 


* 


228 ARITHMETIC. XXXVIII 


44. United States on Russia. Reduce 8672 rouble: 
75 copecks to Federal money; exchange at 32 cents pel 
rouble. 


PRUSSIA. : | 


In Prussia accounts are generally kept in thalers ol 
rix dollars, good gresehen, and pfenings. 

12 pfenings =1 good grosehen, 
24 good evosehen = 1 rix doth) 

The Prussian rix dollar is in value % of the dollar | 

the United States. 

45. Uuited States on Beets, Reduce 4162 rix dok 
lars 18 good groschen to Federal money; exchange at 6€ 
cents per rix dollar. 

46. Prussia on United States. Reduce 3148 dollars 
32 cents to Prussian money; exchange at 1 rix dollar | 
64 cents. 

47. United States on Prussia. Reduce 1428 rix dok 
lars 14 good grosehen to Federal money; exchange at 67 
och per rix dollar. | 

Prussia on United States. Reduce 2136 dollars 
Pederal money to Prussian money; exchange at 1 rix dol- 
lar for 48 cents. ) 


DENMARK. 


In 1813 a new monetary system was established in 
Denmark, in which system the rigsbank dollar is the 
money unit. ‘The denominations of money are the same 
as in the old, or Hamburgh system, but of only half the 
value. 

12 pfenings =1 skilling, 
16 skillmgs ==1 mark, 
6 marks =1 rigsbank dollar. 

The Danish rigsbank dollar is equal to 50 cents in the 
United States. | 

49. United States on Denmark. Reduce 3214 rigs 
bank dollars 4 marks 8 skillings to Federal money; ex 
change at 50 cents per rigsbank dollar. 


| 
: 
| 


| XXXVI. EXCHANGE. 229 
| 50. Denmark on the United States. Reduce 2082 
dollars 35 cents Federal money to Danish money; ex- 
“change at 2 rigsbank dollars per dollar. 

51. United States on Denmark. Reduce 1968 rigs- 
bank dollars 5 marks 12 skillings to Federal money; 

exchange at 48 cents per rigsbank dollar. 

_. 52. Denmark on United States. Reduce 3007 dollars 
Federal money to Danish money; exchange at 2 rigsbank 
dollars 2 skillmgs per dollar. 


| 
| NAPLES. 


_ In Naples accounts are kept in ducati, carlini, and 
grani. The ducat is the money unit, and is divided into 
10 carlins, each of 10 grams, and, by the public banks, 
into 5 tarins of 20 grains each, making ‘he ducat always 
100 grains. 
10 grani 1 carlino, 
10 carlini=1 ducato. 
The value of the silver ducat, in Federal money, is 80 
cents. . 
_ 53. United States on Naples. Reduce 4022 ducati 8 
-carlini to Federal money; exchange at 80 cents per ducat. 
54. Naples on United States. Reduce 1835 dollars 
73 cents Federal money to Neapolitan money; exchange 
at 1 ducato for 78 cents. 3 
55. United States on Naples. Reduce 3508 ducats 
5 carlini to Federal money; at 82 cents per ducat. 
56. Naples on the United States. Reduce 1817 dol 
lars 54 cents I’ederal money to Neapolitan money; ex 
change at 1 ducat for 76 cents. 


SICILY. 


In Sicily accounts are kept in oncie, tari, and grani. 
20 grani= I taro, 


30 tari ==1 oncia. 
Accounts are also kept in scudi, tari, and grani. 
12 tari =1 scudo, or Sicilian crown, 


5 scudi—2 oncie. 
20 


230 ARITHMETIC. XXXVI 
It must be observed, that the denominations of Siciliar’ 
money have but half the value of the same denomination; 
in Naples. The Sicilian oncia, for mstance, passes it 
Naples for only 15 tari, the Sicilian scudo for 6 tari, anc 
other denominations in the same proportion. : 
The scudo is equal to 96 cents, and the oncia to $2.4(| 
Federal money. | 
57. United States on Sicily. Reduce 1214 oncie 20| 
tari 10 grani to Federal money; exchange at 8 cents per 
taro. | 
58. Sicily on United States. Reduce 1457 dollars| 
62 cents Federal money to Sicilian money; exchange at 
1 taro for 8 cents. | : 
59. United States on Sicily. Reduce 3010 scudi 9 
tari 15 grani to Federal money; exchange at 96 cents per 
scudo. 
60. Sicily on United States. Reduce 983 dollars 44 
cents to Sicilian money; enchange at 1 scudo for 95 
cents. 


~ LEGHORN. 


In Leghorn accounts are kept in pezze, soldi, and 


denari di pezza. | 
12 denari di pezza—1 soldo, | 
20 soldi di pezza —1 pezza of 8 reals. 

The value of the pezza is 90 cents, Federal money. 

61. United States on Leghorn. Reduce 2146 pezze’ 
16 soldi 8 denari to Federal money; exchange at 92 cents 
per pezza of 8 reals. 

62. Leghorn on United States. Reduce 1620 dollars: 
45 cents to money of Leghorn; exchange at 1 pezza of 
8 reals for 90 cents. | 

63. United States on Leghorn. Reduce 3293 pezze 
13 soldi 4 denari to Federal money; exchange at 93 cents 
per pezza of 8 reals. | 

64. Leghorn on United States. Reduce 1214 dollars” 
68 cents to money of Leghorn; exchange at 1 pezza of 
8 reals for 91 cents. 


XXXVIII. EXCHANGE. 231 


GENOA. 


In Genoa accounts are kept in lire, soldi, and denari 
di lira; or in pezze, soldi, and denari di pezza; all in 
money fuori banco, or current money. 

12 denari—1 soldo, 
20 soldi —1 lira; 

12 denari—1 soldo, 
20 soldi —1 pezza; 

The value of the pezza is to that of the lira as 4 to 23; 
that is, 4 pezze==-23 lire; therefore 4 soldi di pezza= 
23 soldi di lira; 4 denari di pezza= 23 denari di lira. 

The par value of the lira is 154 cents, that of the pezza 
89 cents, U. S. 

65. United States on Genoa. Reduce 5254 lire 16 
soldi 3 denari to Federal money; exchange at 16 cents 
per lira. 

66. Genoa on the United States. Reduce 1532 dol- 
lars 30 cents to money of Genoa; exchange at 1 lira for 
15 cents. 

67. United States on Genoa. Reduce 8792 lira to 
Federal money; exchange at 164% per lira; | 
68. Genoa on the United States. Reduce 2000 dol- 
lars to money of Genoa; exchange at J lira for 15 cents. 


VENICE. 


In Venice accounts were formerly kept, and exchanges 
computed in ducats, lire, soldi, and denari, moneta pic- 
cola. 


12 denari = 1 soldo, 

20 soldi = 1 lira piccola, 
64 lire piccole= 1 ducat current, 
8 lire piccole. =1 ducat effective. 


The par of the lira piccola, in Federal money, is 9% 
cents. 
Accounts are now kept, and exchanges computed in 
lire Italiane and centimes. 
100 centesimi=— | lira Italiana. 
The common estimate of this money is, that 100 lire 


232 ARITHMETIC. XXXVIIL.- 


piccole are equal to 514 lire Italiane. The lira Italiana 
is of the same value with the French franc. 
69. United States on Venice. Reduce 14642 lire 4 


soldi 8 denari piccoli to Federal money; exchange at 9 


cents per lira piccola. = __ 

70. Venice on the United States. Reduce 814 dollars 
55 cents to Venetian money; exchange at 1! lira piccola 
for 8 cents. : 

We have given the two preceding examples in the old 
currency, for the sake of practice, although it has general- 
ly gone out of use. 

71. United States on Venice. Reduce 6784 lire 
Italiane 40 centimes to Federal money; exchange at 184 
cents per lira Italiana. | 

72. Venice on the United States. Reduce’ 1817 dol- 
lars 82 cents to Venetian money; exchange at 1 lira Ita- 
liana for 18 cents. 7 

73. United States on Venice. Reduce 5236 lire 
Italiane to Federal money; exchange at 183 cents per lira 
Italiana. 


TRIESTE. 


In Trieste accounts are kept and exchanges computed 
in florins and creutzers; or in rix dollars and creutzers. 
4 pfenings = 1 creutzer, 
60 creutzers 1 florm or gulden, 
1 florin, or 90 creutzers=— 1 rix dollar of account. 

The rix dollar specie is equal to 2 florins. 

The par of the florin is 48 cents, in Federal money, 
which makes the dollar of account 72 cents, and the specie 
dollar 96 cents. 

74. United States on Trieste. Reduce 2846 florins 
25 creutzers to Federal money; exchange at 48 cents per 
florin. | 

75. Trieste on United States. Reduce 1637 dollars 
46 cents to money of Trieste; at 1 florin for 47 cents. 

76. United States on Trieste. Reduce 2055 rix dol- 
lars, 25 creutzers to Federal money; exchange at 72 cents 
per rix dollar. 


) XXXVIIL EX CHANGE. 233 


77. Trieste on United States. Reduce 1738 dollars 
88 cents to money of Trieste; exchange at 1 rix dollar 
for 70 cents. 


ROME. 


In Rome accounts are kept, by the old system, in 
scudi, paoli, and bajocchi; quattrini and mezzi quattrini 

are also sometimes reckoned. 

| 2 mezzi quattrini = 1 quattrino, 

5 quattrini == 1 bajoccho; 
10 bajocchi =1 paolo, 

10 paoli, or 100 bajocchi =1 scudo, or Roman crown. 
The Roman crown is equal to the Federal dollar. 
The scudo di stampa d’oro, or gold crown, is equal to 

$1.53. | | 

- When the exchange between the United States and 

- Rome is at par, no reduction is required; for any number 

of scudi and bajocchi are equal to the same number of 

dollars and cents, and the reverse; for mstance, 125 

scudi 75 bajocchi are equal to 125 dollars 75 cents. 

78. Rome on the United States. Reduce 1871 dol- 

lars 19 cents to Roman money; exchange at 1 scudo 2 

bajocchi per dollar. 

‘79. United States on Rome. Reduce 2070 scudi 50 

bajocchi to Federal money; exchange at 101 cents per 

_ scudo. | 

In 1809, the French moneys of account were introduc- 

edinto Rome. ‘The scudo was reckoned at 5 francs 35 

centimes; the franc, therefore, was valued at 18 bajocchi 

3.45 quattrini. 


MALTA. 


Acounts are kept in this island in scudi, tari, and grani, 
20 grani= I taro, 
12 tari ==1 scudo. 
The taro is likewise divided into 2 carlini, and a carline 
“into 60 piccioli. The pezza, or dollar of exchange, is 
equal to 24 scudi. 
90* 1 


934 ARITHMETIC. XXXVIII. 


The par value of the Maltese scudo is 40 cents in 
Federal money. 

The coins in circulation are chiefly Spanish dollars 
and doubloons, and Sicilian dollars and ounces. 'They 
are valued each at a certain rate, as follows, on which a 
variable agio is charged.. 


Spanish dollar 30 tari 10 grani. 
Spanish doubloon = 38 seudi 9 tari. 
Sicilian dollar 830 tari. 

Sicilian ounce ==6 scudi 38 tari. 


80.-United States on Malta. Reduce 1108 Maltese 
scudi 9 tari to Federal money; exchange at 40 cents per 
scudo. 

81. Malta on the United States. Reduce 874 dollars 
76 cents to Maltese money; at 1 scudo for 38 cents. 

82. United States-on Malta. Reduce 3964 Maltese 
scudi 6 tari to Federal money; at 41 cents per scudo. 

83. Malta on the United States. Reduce 674 dollars 
60 cents to Maltese money; at 1 scudo for 40 cents. 


SMYRNA. 


Accounts are kept here in piastres or gooroosh. The 
piastre, also called the Turkish dollar, is divided some- 
times into 12 temins, sometimes into 40 paras or medini; 
but the usual division is into aspers, the number of which 
varies. Thus, the English and Swedes divide the piastre 
into 80 aspers; the Dutch, French, and Venetians into 
100 aspers; the Turks, Greeks, Persians, and Armenians 
into 120 aspers. An asper is a third part of a para. 

Bills o. exchange are often drawn on Smyrna in foreign 
coin, particularly in Spanish dollars, which are always to 
be had there; but, if drawn in a com not in current use, 
the exchange of the day is established to make the pay- 
ment. : 

The Turkish coins, owing to the frequent deterioration 
of them by the government, have been declining in their 
intrmsic worth for many years, and have no standard 
value. Foreign exchanges are conducted entirely ac- 
cording to the price of the day. : 


Fd 


XXXVIII. EXCHANGE. 235 


84. United States on Smyrna. Reduce 53183 piastres 
of Turkey to Federal money; exchange at 20 cents per 
piastre. . 

85. Smyrna on the United States. Reduce 912 dol- 
lars 27 cents to Turkish money; exchange at 1 piastre 
for 21 cents. 

86. United States on Smyrna. Reduce 71614 Turk- 
ish piastres to Federal money; exchange at 22 cents per 
piastre. 

87. Smyrna on United States. Reduce 1128 dollars 
22 cents to money of Smyrna; exchange at 1 piastre for 
204 cents. 


EAST INDIES. 


Before European colonies were established in the East 

Indies, particularly while the power of the Moguls pre- 
vailed in Hindostan, the monetary system was very simple. 
There was current throughout these vast dominions one 
principal coin of silver, denominated the sicca rupee. It 
was of a certain weight called the sicca. The siccu was 
‘used also as a standard for weighing other articles. 
_. The British possessions in the. Kast Indies are divided 
‘into three presidencies, viz. Bengal, Bombay, and Ma- 
dras. The monetary systems in these presidencies are 
different from each other. 


CALCUTTA IN BENGAL. 


Accounts are commonly kept here in current rupees, 

annas, and pice. 
12 pice = I anna, 
16 annas= 1 rupee, currency. 

The East India Company, however, keep their ac- 
counts in sicca rupees, similarly divided, which bear a 
batta or premium of 16 per cent. above current rupees. 

The current rupee of Calcutta is 447% cents, and the 
sicca rupee 51-7, cents, in Federal money. 

A Lac of rupees is 100000, and a Crore of rupees is 
100 Lacs, or 10 millions of rupees. 


236 ARITHMETIC. XXXVI 


88. United States on Calcutta. Reduce 17438 rupees: 
12 annas, currency of Calcutta, to Federal money; ex- 
change at 48 cents per rupee. 

89. Calcutta on the United States. Reduce 6913 
dollars 25 cents to money of Calcutta; exchange at 2 
sicca rupees per dollar. et | 

90. United States on Calcutta. Reduce 46173 cur- | 
rent rupees 9 annas to Federal money; exchange at 46. 
cents per rupee. 

91. Calcuttaon United States. Reduce 28953 dollars | 
63 cents to current money of Calcutta; exchange at 1 
rupee for 44 cents. | 

92. United States on Calcutta. Reduce a Lac of 
sicca rupees to Federal money; exchange at 53 cents per 
sicca rupee. 


BOMBAY. 


In the presidency of Bombay, accounts are kept in 
rupees, quarters, and reas. f 
100 reas —=1 quarter, © 

| 4 quarters—1 rupee. : 

The current value of the Bombay rupee is equal to 50 
cents in Federal money. | i 

93. United States on Bombay. Reduce 10137 rupees 
~ quarters 50 reas to Federal money; exchange at 50 
cents per rupee. ' 

94. Bombay on the United States. Reduce 6210- 
dollars 48 cents to money of Bombay; exchange at 48 
cents per rupee. 

95. United States on Bombay. Reduce 8413 rupees 
of Bombay to Federal money; at 49 cents per rupee. 


MADRAS. 


In the presidency of Madras, there are different mone- 
tary systems, which may be distinguished under the heads 
of the old system and the new. , 

According to the old system, accounts are kept in star 
_ pagodas, fanams, and cash. * 


XXXVIII. EXCHANGE. 237 


80 cash ==1 fanam, 
42 fanams = 1 pagoda. 
The current value of the star pagodais $1.80 Federal. 
By the new system, the silver rupee of Madras is made 
the standard coin, and money of account in this presidency. 
The current value of the silver rupee of Madras is 
44,7 cents. It is divided into halves, quarters, eighths, 
and sixteenths. The sixteenth is the anna. 
96. United Stateson Madras. Reduce 72183 rupees 
of Madras to Federal money; at 46 cents per rupee. 
97. Madras on United States. Reduce 2684 dollars 
85 cents to money of Madras; at 44 cents per rupee. 
98. United States on Madras. Reduce 5367 rupees 
of Madras to Federal money; at 45 cents per rupee. 


CANTON IN CHINA. 


In China accounts are kept in tales, mace, candarines, 
and cash. 


10 cash | ==1 candarine, 
10 candarines= 1 mace, 
10 mace ==} tale. 


The tale is reckoned at $1.48 in Federal money. 

99. United States on Canton. Reduce 12144 tales 
5 mace to Federal money; exchange at 1 dollar 48 cents 
per tale. 

- 100. Canton on the United States. Reduce 8754 
dollars 89 cents to money of Canton; exchange at 1 tale 
per 146 cents. 

- 101. United States on Canton. Reduce 16235 tale 
to Federal money; exchange at 149 cents per tale. 


JAPAN. 


In the empire of Japan, which consists of severa: 
islands to the east of Asia, accounts are kept in tales. 
‘mace, and candarines. 


10 candarines= 1 mace, 
10 mace = I tale. 
The Japanese tale is reckoned at 75 cts, Fed. money 


| 


: 


238 ARITHMETIC. XXXVIIL. 


102. United States on Japan. Reduce 3714 Japan- 
ese tales 8 mace 8 candarines to Federal money; exchange 
at -75 cents per tale. | 

103. Japan on United States. Reduce 696 dollars 54 
cents to money of Japan; exchange at | tale per 75 cents. 

104. United States on Japan. Reduce 2468 tales 5 
mace, money of Japan, to Federal money; exchange at 76 
cents per tale. 


SUMATRA. 


This island is chiefly in possession of the natives; but 
the English have a small settlement at Bencoolen. 
At Bencoolen accounts are kept in dollars, soocoos, 
and satellers. 
8 satellers—1 soocoo, 
4 soocoos —1 dollar. 
This dollar is reckoned at $1.10 in Federal money, 
and is sometimes called a rial. 4 
105. United States on Bencoolen. Reduce 1947 
Bencoolen dollars 3 soocoos 4 satellers to Federal money; 
exchange at 110 cents per dollar of Bencoolen. 
106. Bencoolen on United States. Reduce $2379.51 
Federal money to money of Bencoolen; exchange at 1 
dollar Bencoolen for 108 cents. 


ACHEEN. ( In the island of Sumatra). 


In Acheen accounts are kept in tales, pardows, mace, 
and copangs. 
4 copangs 1 mace, 
4 mace =1 pardow, 
4 pardows = | tale. 
The mace is a small gold coin worth about 26 cents’ 
Federal money, which makes the tale $4.16. bs 
107. United States on Acheen. Reduce 1432 tales, 
3 pardows 2 mace to Federal money; exchange at 416 
cents per tale. . 
108. Acheen on United States. Reduce 3620 dollars — 
964 cents to money of Acheen; at 1 tale for 412 cents. — 


oa 


® 


" 
i 


XXX VIII. EXCHANGE. 239 


JAVA. 


In Batavia, the capital of this island, the florin or guil- 
der of the Netherlands is the monetary unit; but instead 
of the decimal divisions, it is here sometimes divided into 
schillings, dubbels, stivers, and dotts. 

5 doits = 1 stiver, . 

2 stivers ==1 dubbel, 

3 dubbels =1 schilling, 

4 schilling=1 florin or guilder. 

The florin of Java, as the florin of the Netherlands, is 
equal to 40 cents Federal money. 

109. United States on Batavia. Reduce 11841 florins 
3 schillings 2 dubbels to Federal money; exchange at 40 
cents per florin. 

110. Batavia on the United States. Reduce $ 13746. 
69 to money of Batavia; exchange at 1 guilder for 38 
cents. 

111. United States on Batavia. Reduce 42328 guil- 
ders 50 centimes to Federal money; exchange at 42 cents 
per guilder. | 


MANILLA. (In the island of Luzon). 


In Manilla, the capital of the Spanish East India pos- 
sessions, accounts are kept in Spanish dollars or pesos, 
reals, and maravedis. - 

34 maravedis —1 real, 
8 reals = 1 dollar. 

112. United States on Manilla. Reduce 6341 dollars — 
6 reals 17 maravedis to Federal money; exchange at 101 
cents per Spanish dollar. 

113. Manilla on United States. Reduce $5274.55. 
to money of Manilla; exchange at 1 Spanish dollar per 
dollar. 


COLOMBO. (In the island of Ceylon). 


In Colombo, accounts are kept in ri dollars, fanams 
and pice. 


240 ARITHMETIC. | XXXVIIL| 


4 pice =1 fanam, 

12 fanams = 1 rix dollar. _... | 

The current value of this rix dollar is 40 cts. F. money, 

114. United States on Colombo. Reduce 7328 rix 

dollars 9 fanams to Federal money; exchange at 40 cents 

per rix dollar. 

115. Colombo on United States. Reduce $1426.71 

Federal money to money of Colombo; exchange at 1 rix 
dollar per 38 cents. 


MAURITIUS, (Isle of France.) 


In Mauritius, accounts are kept in two different ways, 
viz. in dollars of 100 cents, which is the mode adopted 
in public or government accounts; and in dollars, livres, 
and sols, which method is mostly used by merchants. 

20'sols =1 livre. 
10 livres= 1 dollar. 

‘These are called colonial livres, and are 10 cents each. 

116. United States on Mauritius. Reduce 4132 dol- 
lars 7 livres 10 sols to Federal money; exchange at 1 
dollar per dollar of Mauritius. 

117. Mauritius on United States. Reduce $7547.47, 
Federal money, to money of Mauritius; exchange at 98 
cents per dollar of Mauritius. 


ARBITRATION OF EXCHANGE. 


ARBITRATION OF EXCHANGE is a comparison of the 
courses of exchange between different countries, in order 
to ascertain the most advantageous course of drawing or 
remitting bills. It is distinguished into simple and com- 
pound arbitration. 

Simple Arbitration is a comparison between the ex- 
changes of two places through a third; that is, it is finding 
such a rate of exchange between two places, as shall be 
in proportion to the rates quoted between each of them 
and a third place. The exchange thus determined is 
called the arbitrated price. 


XXXVIIL. EXCHANGE. 241 


~ If, for example, the course of exchange between Lon- 

don and Paris is 24 francs for 1 pound sterling, and be- 
tween Paris and Amsterdam 54 pence Flemish for 3 
francs, the arbitrated price between London and Amster- 
dam through Paris, is 36 shillings Flemish for 1 pound 
sterling; for, as 3fr. : 24fr.—54d. : 36s. Flem. 

Suppose the arbitrated price to be, as before stated, 

~36s. Flemish for £1 sterling; and suppose the direct 
course between London and Amsterdam to .be 37s. 
Flemish; then London, by drawing directly on Amster- 
dam, must give 37s. Flemish for £1 sterling; whereas, 
by drawing through Paris, he will give only 36s. Flemish, 
for £1 sterling. « It is therefore the interest of London 
to draw indirectly on Amsterdam through Paris. 

On the contrary, if London remits directly to Amsiter- 
dam, London will receive 37s. Flemish for £1 sterling; 
but, by remitting through Paris, London will receive only 
36s. Flemish. It is the interest of London, therefore, 

to remit directly to Amsterdam. 

118. If the exchange of London with Genoa is 47d. 
sterling per pezza, and that of Amsterdam with Genoa 86 
grotes Flemish per pezza, what is the proportional or 

-arbitrated exchange between London and Amsterdam 
through Genoa? that is, how many shillmgs and grotes 
Flemish are equal to £1 sterling ? 

Since 47d. sterling is equal to 1 pezza, and this pezza 
is equal to 86 grotes Flemish, the question may be stated 
thus; 47d. sterling : 240d. sterling =86 grotes Flemish: 
Ans. which is 36s. 77% grotes Fl. By the Chain Rule, 

(See Art. xxvr), the statement is as follows. 
: 1 pound sterling. 
‘1 pound sterling = 240 pence. 


47 pence = | pezza. 
1 pezza = 86 grotes Fl: 
12 grotes = 1 shilling FI. 


The product of the consequents being divided by the 
product of the antecedents, will give 36sh. 77 grotes 
]. for the answer. 
119. If the exchange on London with Hamburgh is 
34 shillings 2 grotes Flemish banco for £1 sterling, and 
21 


242 ARITHMETIC. XXXVIIL. 


_ that of Amsterdam with Hamburgh 333 stivers per rix 
dollar of 2 marks, what is the arbitrated exchange be-— 
tween London and Amsterdam through Hamburgh? 
Since 2 marks are 64 grotes Flemish and 333 stivers 
are 663 grotes Flemish, the question may be stated thus, 
64 grotes Fl.: 662 grotes Fl. = 34s. 2 grotes Fl.: Ans. 
By the Chain Rule, the statement is as follows, 
1 pound sterling. 
1 pound sterling — 34 s. grotes Flem. 


8s. Flem. core (ON WAT K Se 
2 marks == 33% stivers. 
_ 6 stivers = ion Flemish. 


120. If the exchange of London on Leghorn is 514d. 
sterling per pezza, and that of Amsterdam on Leghorn 
923 grotes Flemish per pezza, what is the proportional 
exchange between London and Amsterdam through Leg-— 
horn? : 
121. If the exchange of London on Lisbon be 68d." 
sterling per milree, and that of Amsterdam on Lisbon 48 
grotes Flemish per old crusado, what is the arbitrated ex-- 
change between London and Amsterdam through Lisbon ? 

122. If the exchange of London on Madrid is 42d. 
sterlmg per dollar of plate, and that of Amsterdam on” 
Madrid 96 grotes Flemish per ducat of plate, what is the 
proportional exchange between London and Amsterdam 
through Madrid? | 

123. If the exchange of London on Paris is 24 francs 
per £1 sterling, and that of the United States on Paris 
185 cents per franc, what is the arbitrated or propor- 
tional exchange between London and the United States 
through Paris? 

124. If the exchange of London on Amsterdam is 11 
florins 16 stivers per £ sterling, and that of the United 
States on Amsterdam 38 cents per florin, what is the’ 
arbitrated exchange between the United States and Lon- — 
don through Amsterdam? 

125. Ifthe exchange of the United States on Paris is” 
18 cents per franc, and that of Amsterdam on Paris 547 
grotes “semish for 3 francs, what is the proportional 

e,change between the United States and Amsterdam — 
through Paris? 


XXXVIII. EXCHANGE. . Q43 


126. If the exchange of the United States on Lisbon 
is $1.24 per milree, and that of Paris on Lisbon 540 
rees per ecu of 3 francs, what is the proportional ex- 
change between the United States and Paris through 
Lisbon? 


COMPOUND ARBITRATION. 


Compound arbitration is a comparison between the 
exchanges of more than thrce places, to find the arbitrat- 
ed price between the first place and the last, in order to 
ee on the most advantageous mode of negotiating 

ills. 

127. Suppose the exchange between London and 
Amsterdam to be 35 shillings Flemish for £1 sterling; 
between Amsterdam and Lisbon, 42 pence Flemish per 
old crusado; and between Lisbon and Paris, 480 rees 


per ecu of 3 francs; wnat is the arbitrated price between 


London and Paris? 
First, 35s. Fl. : 42d. Fl. £1 sterling : A; which is 
2s. sicrling. 
Secondly, 1 old crusado : 480 rees==2s. sterling : A; . 
which is 2s. 44d. sterling. 
Thirdly, 2 s.44d, sterling : £1 sterling =3 francs : 
A; which is 25 francs. 
Hence the arbitrated price is 25 francs for £1 sterling. 
But all such operations are best performed by the Chain 
Rule; thus, 1 pound sterling. 
1 pound sterling 36 shillings Flemish. 
33 shillings Fl. 1 old crusado. 
1 old crusado 400 rees. 
480 rees 3 francs. 
The product.of the consequents divided by that of 
the antecedents gives 25 francs per £ sterling, as before. 
128. Suppose a merchant in London has a sum of 
money to receive in Cadiz, the exchange being at 38 d. 
sterling per dollar of plate; but, instead of drawing di- 
rectly on Cadiz, he draws on Amsterdam, ordering his 
agent there to draw on Paris, and Paris to draw on Ca- 
diz; the exchange between London and Amsterdam being 


Al 


4 


' 44 ARITHMETIC. XXXVILI. 


at 35 shillmgs Flemish per pound sterlmg; between Am-— 


sterdam and Paris 534 grotes Flemish per ecu of 3_ 


francs; and between Paris and Cadiz 15 francs 50 cen- 
times per doubloon of plate. What is the arbitrated 
price between London and Cadiz ? 
1 dollar of plate. 
4 dollars of plate == 1 doubloon of plate. 
1 doubloon of plate— 15% francs. + 
3 francs = 531 grotes FI. 
12 grotes FI. == 1 shillng FI. 
35 shillings FI. == 240 pence sterling. 
The result is 3944d. sterling per dollar of plate. 
The circuitous operation is, therefore, the most advanta- 


Fi aaa SOS 


geous, as London gets 393d. nearly, instead of 38d. for é 


each dollar of plate. 


129. London having a sum to receive in Lisbon, when — 
the exchange is at 64d. sterling per milree, draws on © 


Lisbon, but remits his bill to Hamburgh to be negotiated, 
and directs the returns to be made to him im bills on 


Leghorn; the exchange between Hamburgh and Lisbon ~ 


being 45 grotes Flemish per old crusado; between idam- 


burgh and Leghorn 85 grotes Flemish per pezza; and — 


between London and Leghorn 52d. sterling per pezza. 
What is the arbitrated price between London and Lis- 


bon? and what does London give per milree by the cir- — 


culitous exchange ? 
{30. A merchant in London has a sum to pay in 


Petersburg, and another to receive in Genoa; but there — 


being no regular exchange between these places, London 


draws on Hamburgh, and remits his bill to Petersburg, — 


directing Hamburgh to draw on Genoa; the exchange 
between London and Genoa being 463d. sterling per 
pezza; between Hamburgh and Genoa 81 grotes Flemish 
per pezza; and between Petersburg and Hamburgh 23 
schillings Lubs per ruble. What is the exchange between 
London and Peter rsburg resulting from the operation? 


that is, how many pate sterling does London pay for 


the ruble ? 
131. A merchant in the United States has funds in 
Paris, and owes a sum of money in Hamburgh; he draws 


| 
: 
| 
XXXVIII. FOREIGN COINS. 245 


on London, remits his bill to Hamburgh, and directs 
London to draw on-Paris; the exchange between the 
United States and Paris being 18 cents per franc; be- 
tween London and Paris 24 francs 25 centimes per £ 
sterling; and between Hamburgh and London 13} marks 
banco per £ sterling. How many cents per mark banco 
does the American merchant pay by this course of ex- 
change ? 

132. A merchant in the United States being indebted 
in London, remits bills on Paris to his correspondent in 
that city, and directs him to obtain bills of Paris on Lis- 
bon and remit them to his creditor in London; the ex- 
change between the United States and France being 18 
cents per franc; between Paris and Lisbon 465 rees per 
ecu of 3 francs; and between London and Lisbon 63d. 
sterling per milree. In this course of exchange, how 
many pence sterling are paid with one dollar of the Unit- 
ed States ? 

In the preceding examples, no notice is taken of the 
expenses incident to exchange operations, such as com- 
mission, brokerage, interest, &c.; but in all transactions 
of business, it is necessary to make allowance for the 
difference of charges between direct and indirect ex- 
changes, in order to decide on the preference of the one 
to the other. | 


FOREIGN COINS. 


THE SILVER COINS of foreign countries, rendered 
current in the United States, by Act of Congress, are 
_as follows. Spanish dollars and parts thereof, at 100 
cents the dollar. Dollars of Mexico, Peru, Chili, and 
Central America, of not less weight than 415 grains each, 
and those restamped in Brazil of the like weight, and of 
not less fineness than 10 oz. 15 dwt. pure silver in the 
Troy pound, all at 100 cents the dollar. The Five- 
franc pieces of France, weighing 384 grains each, and of 
not less fineness than 10 oz. 16 dwt. pure silver in the 
Troy pound, at 93 cents the piece. 
21* 


246 ARITHMETIC. XXXVII1. 


THE GOLD COINS of foreign countries, with their 
respective weights, and values, are stated in the following © 
TasLe. Those of the countries printed in Italics are 
rendered current, by Act of Congress. 


Weight. Value. 


Names of Countries and Coins. - dwt. gr. | $ cts m 


AUSTRIAN DOMINIONS. 


(oor) 


Ie > 


Souveremowye Ce Vise Gh Pale 3 14 3 37 7 

Double Dudaty. 82h ae! Pe ee See @ 

Hungarian, Ducat, 2 53) 2296 
BAVARIA 

Carolin, . Te TRA. 6 54] 495 7 

Max d’or, or Maximilian, mes 4°40 S8Sr’s 

Ducat, zee WP ath, 2 53) 2275 
BERNE. a 

Ducat, double in proportion oe 1-93 (1 1°98 

Pistole. Set ees od 4 21 | 4 54 
BRAZIL. 

Jahannes, half in proportion. . | 18 17 06 

Doviraon, © ss Pees aa eS uae 

PUA. ks OEE Ga ee 

Moidore, half in | proportion, a ds a 

Crusade, .. Bes 164 63 5 
BRUNSWICK. 

Pistole, double in atin ; 4 211} 4548 

Diva ors eM as B $3) 2126 
COLOGNE. 

Ducat ae Oi OPA Pe ice, 25352 Borg 
COLOMBIA. 

Doublbon! tgp eae ke 17° 6" (1S S35 
DENMARK. 

Ducat, Current, a 2 1812 

Dileat"Specie;s ois <0 2° 53-2 26M) 

Christian d’or, 4 4 02 1 
EAST INDIA. 

Rupee, Bombay, 1818, TAL. [rooms 

Rupee, Madras, 1818, a) AD eee 

Pagoda, Star, 2 42) 1 7958 
ENGLAND. 

Guinea, half in proportion, 5 81| 5 07 5 

Sovereign, half in proportion, . 5 24| 4 846 


XXXVIIL. 


Seven Shilling Piece, 

FRANCE. 

Louis, coined before 1786, 
Double Louis, before 1786, . 
Louis, coined since 1786, 
Double Louis, since 1786, 
Napoleon, or 20 francs, 
Double Napoleon or 40 francs, . 
Same as the new Louis Guinea, 

FRANKFORT ON THE MAIN. 

Ducat, Shee 

GENEVA. ¥ 
Pistole, old, 

Pistole, new, 

HAMBURG. 

Ducat, double in proportion, 

GENOA. 

Sequin, . 

HANOVER. 

- Double George d’or, ae in Ng 
Ducat, . 
Gold Florin: double! in pro a 

HOLLAND. 

Double Ryder, 

Ryder, , 

Poratss ss *: 

Ten Guilder Piece, 5 do. in n pro’n ‘n, 

MALTA, 

Double Louis, . 
Louis, 
Demi Louis, 

MEXICO. 

Doubloons, shares in pro’n, . 

MILAN. 

Sequin, . . 
Doppia or Pistole, 
Forty Livre Pieces, 1808, 

NAPLES. 

Six Ducat Piece, 1783, . 
Two do. or Sequin, 1762, 
Three do. or Oncetta, 1818, 


FOREIGN COINS. 


posed, 


eat PS) 
on on os Or “IO © 
' Blo Alo Pwd Alo die 


— 


to BNO me OP 


oo 09 


onal 


s% = 


248 ARITHMETIC. XXXVIIL. 
NETHERLANDS. 


Gold Lion, or 14 Florin ia 5 731 5 04 6 
Ten Florin Piece, 1820,.  . 4 73|.4 019 
PARMA. 
Quadruple Pistole, double in pro’n, 18.59. 16,632.85 
Pistole or Droppia, 1787, : 414 ;,4194 
Pistole or Droppia, 1796, bon 414 |} 4135 
Maria Theresa, 1818, vis 4 33] 3 86 1 
PIEDMONT. 
Pistole c’d since 1785, 3 10 pro’n, 5 20 | 5.41.0 
Sequin, haif in proportion, 2) .5. | sae 
Carlino, c’dsince1785,4inpro’n, |29 6 |27 34 
Piece of 20 Francs, or Marengo, 4 31) 3 564 
POLAND. - 
Ducat Seek ss sa le ean 
PORTUGAL, | 
Dobraons Py. a ce) eh ol ee 
Dobra, toe 530.) oa ot & «Be 
Johannes, . .- eS 17 06 4 
Moidore, half in proportion, aa 6.22 6.58 7 
Piece of 16 Testoons, 1600 Rees, 2 6 |; 2.12.19 
Old Crusado or 400 Rees, . . 15) bs pesig 
New Crusado or 480 Rees,. . 164 63 5 
Milree, coined in 1755, . 193 73 
PRUSSIA. | 
WoC Nei, oka koi ea ge 53 
Ducat, 1787, 2 52 
Frederick, dauble, 1769, mass 8 14 
Frederick, double, 1800, ae: 8 14 
Frederick, single, 1878, pL, 
Frederick, single, 1800, 4 7 
ROME. 
Sequin, coined since 1760, : 2 44 
Scudo of Republic, Se Wied) 8 
RUSSIA. 
PUA Es Wanye a 2.26 
Ducat, 1763;-4. : 2° 63 
Gold Ruble, 1756, 01 
Gold Ruble, 1799, 183 
Gold Polten, 1777, 9 
Imperial, 1801, ewe 


XXXVI. 


’ Half Imperial, 1801, 

- Half Imperial, 1818, 

SARDINIA. 

Carlino, half in proportion, 

SAXONY. 

Ducat, 1784, 
Ducat, 1797, 
Augustus, 1754, 
Augustus, 1784, 

SICILY. 

Ounce, 1751, . 
Double Ounce, 1758,. 

SPAIN. 

Doubloons, 1772, double and sin- 
gle and shares in proportion, . 
Doubloon, Ay. dew 
Pistole, .. 

Coronilla, (Gold Dol. a or Vintern, 1801, 

SWEDEN, 

Ducat, 

SWITZERLAND. 

Pistole of Helvetic Republic, 1800, 

TREVES. 

Ducat, 

TURKEY. 

+ Sequin fonducli, of Cons’ple, 1773, 
Sequin fonducli, of Cons’ Be 1789, 
Half Misseir, 1818, 

Sequin Fonducli, 
Yeermeeblekblek, 

TUSCANY. 

Zechino, or Sequin, 
Ruspone of kingdom of Etruria, 
VENICE. 


b] 


Zechino, or Sequin, shares in pro. 


WIRTEMBURG. 
Carolin, . 
Ducat, 
ZURICH. 
Ducat, double and half in pro’n, 


FOREIGN COINS. 


—t 
ret Ot CO Or Cr 


Ivo 


—" 


On 
Halo 


“Et On 


DB) Dla 


WD] AI co|co 


vl 


249 


G9 09 & WH © oo 09 
LS) 
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moron iS) €9 0 


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250 ARITHMETIC. XXXVIII. 


FOREIGN WEIGHTS AND MEASURES. 


The weights and measures of GREAT BRITAIN are 
the same as those of the United States, excepting the 
variations which are noted in the tables of “ Weights and 
Measures,’ page 27. 


The weights and measures of FRANCE being more 


nicely adjusted than those of any other country, will be 


here given the more fully on that account. It is, however, | 
to be observed, that these weights and measures are ac- 


cording to a new system, not yet in very common use. 


The fundamental standard adopted in France for the 
metrical system of weights and measures, is a quadrant 
of the meridian; that is to say, the distance from the. 


equator to the north pole. This quadrant is divided into 


ten millions of equal parts, and ones of these equal parts” 
is called the Metre, which is adopted as the unit o1 
length, and from which by decimal multiplication and 


division all other measures are derived. 


In order to express the decimal proportions, the fol-' 


lowing vocabulary of names has been adopted. 
Fot multipliers, 


the word Deca prefixed, means - 10 times.” 
re dd eetoe ©" e 100 times. 
ee Chilo eo J 1000 times. 
eye © | hy _ 10000 times 


For divisors, 


the word Dect prefixed, expresses the 10th part. | 


ae enti. 8° eS — 100th part. 


“Milli « “ 1000th part. 


It may assist the memory to observe that the terms for 
multiplying are Greek, and those for dividing, Latin. - 
Thus, Deca-metre means 10 Metres. 
Deci-metre ‘* the 10th part of a Metre. 


Flecto-metre ‘* 100 Metres. cok 


Centi-metre ‘‘ the 100th part ofa Metre; &c. 


= 


XXXV11II. WEIGHTS AND MEASURES. 251 . 


Frencu Lona MEASURE. 


The Metre, which is the unit of long measure, is equal 
to 39.371 English inches. 


10 milli-metres ==1 centi-metre, 
10 centi-metres =1 deci-metre, 

10 deci-metres =—=1 METRE 

10 Metres. . ==1 deca-metre, 

10 deca-metres 1 hecto-metre, 
10 hecto-metres —1 chilo-metre, 
10 chilo-metres —1 myria-metre. 


Frencu Square MEASURE. 


The Are, which is a square deca-metre (or 100 square 
Metres), is the unit of square or superficial measure, and 
is equal to 3.953 English square rods. 


10 milliares . . ==1 centiare; 
10 centiares . . ==1 deciare; 
10 deciares . . ==1 ARE; 
10 Ares . «© sz=l1 decare; 
10 decares . . =I hectare; 
10 hectares . . = 1 chilare; 
10 chilares . . ==1 myriare. 


Frencu MEASURES OF Capacity. 

The Litre, which is the cube of a decimetre, is the 
unit of all liquid measures, and of all other measures of 
eapacity. The Litre is equal to 61.028 English cubic 
nches. 


10 millilitres. . == 1 centilitre; 
10 centilitres. . ==1 decilitre, 
10 decilitres . . ==! LitRE; 
10 Litres . . =1 decalitre; 
10 decalitres. . 1 hectolitre; 
10 hectolitres . ==1 chilolitre; 
10 chilolitres. . ==1 myrialitre. 


Frencu Sotip MEASURE. 

The Stere, which is a cube of the metre, is the unit of 
solid measure, that is used for fire-wood, stone, &c. 
The Stere is equal to 35.31714 English cubic feet; it is 
the same as the chilolitre in measures of capacity. 


252 “ARITHMETIC. XXXVIIL 


10 decisteres. 1 Srerg; 
10 Steres. . ==1 decastere. 


Frencu Werents. 
The Gramme, which 1s the weight of a cubic centi« 
metre of distilled water of the temperature of melting 
ice, is the unit of all weights. The Gramme is equal to 


15.434 grains Troy. Grains Troy 

A milligramme is 1000th part of a gramme, = 0.0154 

A centigramme is 100th part of a gramme,== 0.1543 
A decigramme is 10th part of a gramme, = 1.5434 
-A GRAMME ees 15.4340 

A decagramme is 10 grammes, = 154.3406) 
A hectogramme is 100 grammes, = 1543.4000 


Achilogramme is 1000 grammes, 15434.0000 | 
A myriagramme is 10000 grammes, == 154840.0000 


~ All the preceding French weighis and measures are de- 
duced from some decimal proportion of the metre. ‘Thus 
the chilogramme corresponds with the contents of a 
cubic vessel of pure water at the lowest temperature, the 
side of which vessel is the tenth part of the metre (the 
decimetre), and the gramme answers to the like contents 
ofa cubic vessel, the side of which is the hundredth part 
of the metre (the centimetre); for the contents of all 
cubic vessels are to each other in the triplicate ratio of 4 
their sides. 


100 lb. of HAMBURGH 106.8 jb. avoirdupois. 


The shipfund is 280 Ib. = 299 Ib. avoirdupois. 
1 foot, Hamburgh = 11.289 inches, U. S. 
The Hamburgh ell is 2 feet ==22.578 inches, U.S. 
The Hamburgh mile = 4.684 miles, U.S. 
The fass of Hamburgh = 1.494 bushel of U.S. 
The last of grain is 60 fasses —89.64 bushelsof U.S. 
The ahm of Hamburgh = 38.25 gallons, U. S. 


100 Ib. of AMSTERDAM 108.93 Ib. avoirdupois. 
4 shipfunds is 1 ship-pound ==326.79 Ib. avoirdupois. — 
The Amsterdam last == 85.248 bushels, U.S. 


KXXVIII. WEIGHTS AND MEASURES. 253 


The Aam (liquid) ==41 gallons, U. States. 
The Amsterdam foot = 11.147 inches, U. S. 
The ell of Amsterdam = 27.0797 inches, U.S. 
The ell of the Hague == 27.333 inches, U. S. 
The ell of Brabant == 27.585 inches, U. S. 
100lb. of PORTUGAL = 101.19 lb. avoirdupois. 
An arroba is 32 lb. = 32.38 lb. avoirdupois. 
The moyo, a dry measure ==23.03 bushels, U. 8. 


~The almude, a liquid measure 4.37 gallons, U. 5. 
The pe or foot, long measure 12.944 inches, U. 8. 
The palmo or standard span ==8.64 inches, U. 8. 


The vara is 5 palmos -~)==43.2 Mehes Uses. 
The Portuguese mile =21)26Cnilep Uie8. 
100 lb. of SPAIN == 101.44 lb. avoirdupois. 
The arroba of wine == 4,245 gallons, U. 8S. 
The fanega, 75 of a cahiz 1.599 bushels, U. S. 
The Spanish standard foot = 11.128 inches, U. S. 
The vara, a cloth measure —= 33.384 inches, U.S. 
The legua or league 4.291 miles, U. S. 
100|b. victualie, of SWEDEN = 93.76 |b. avoirdupois. 
The Swedish foot = 11.684 inches, U. S. 
The Swedish ell is 2 feet —= 23.368 inches, U. S. 
~The Swedish mile — 6.64 miles, U. S. 
- The kann, (both dry and liquid) =1593 cubic in. U. 5S. 
100 kanns 69.09 galls. wine,U.S. 
100 kanns —-7.42 bushels, U. S. 
100 lb. of RUSSIA == 90.26 lb. avoirdupois. 
400 lb. make 1 berquit == 361.04 lb. avoirdupois. - 
A pood is 40 lb. Russian == 36.1054 Ib. avoir’s. 
A chetwert, a dry measure, ==5.952 bushels, U.S. 
The vedro, a liquid measure, ==3.246 gallons, Ue 
The Russian inch —=1 inch, U. S. 
_ The Russian foot —13.75 inches, U. S. 
- The arsheen, a cloth measure, = 28 inches, U. 5. 
The sashine or fathom = 7 feet, U. S. 
A werst or Russian mile == 3500 feet, U. S. 


22 


254 ARITHMETIC. XXXVIII 


100Ib. of PRUSSIA = 103.11 Ib. avoirdupois 
The quintal is 110 lb. = 113.421 lb. avoir’s. 

The scheffel, a dry measure, ==1.5594 bushel, U. S. 
The eimer, a liquid measure, =—=18.14 gallons, U. S. : 


The Prussian foot == 12.356 inches,.\U.493 
The Prussian ell = 26.256 inches, U. S. | 
The Prussian mile ==4..68 miles s1Us Site | 
100 lb. DENMARK, = 110.28 lb. avoir’s. | 
The centner is 100 lb. ==110.28lb. avoir’s. | 
The shippond is 320 lb. = 352.896 lb. 


The bbl.or toende, a dry meas. 3.9472 bushels, U. S. 
The viertel, a liquid measure. =2.041 gallons, U. S. 
The Danish or Rhineland foot 12.356 inches, U. S. 


The Danish ell is 2 feet == 24.712 inches, U. S. 
The Danish mile = 4.684 miles, U. S. 

A cantaro grosso, NAPLES, =196.5 lb. avoirdupois. 
The cantaro piccolo = 106 lb. avoirdupois. 
The tomolo, a dry measure, ==1.451 bushels, U. S. 
The carro is 36 tomoli -==52.236 bushels, U.S. 


The barile, a liquid measure, ==11 gallons. U.S. 
The carro of wine is 24 barili 264 gallons. U. S. 


The palmo, long measure, —10.38 inches, U. S. 
The canna is 8 palmi = 83.04 onen U. S. 
100 1b. or libras, SICILY, |= =70|b. avoirdupois. 
The cantaro grosso = 192.5 lb. avoirdupois. 
The cantaro sottile = 175 lb. avoirdupois. 
The salma grossa, adry measure, 9.77 bushels, U. S. 
The salma generale == '7385 bushels, U.S. 
The salma, a liquid measure, ==23.06 gallons, U. S. 
The palmo, a long measure, =9.5 inches, U. S. 
The canna is 8 palmi == 76 inches, U. S. 
100 lb. of LEGHORN, —=75 lb. avoirdupois. 
The sacco, a dry measure, ==2-', bushels, U. S. 
The barile, a liquid measure, ==12 gallons, U. S. 
155 braccia, cloth measure, © ==100 yards, U. S. 


The canna of 4 braccia == 93 inches, U. S. 


XXXVI. WEIGHTS AND MEASURES. 200 
100 lb. peso grosso of GENOA, = 76.875 lb. avoir’s. 


100 lb. peso sottile == 69.89 lb. avoir’s. 
The mina, a dry measure, == 3.426 bushels, U. S. 
‘The mezzarola, liquid measure, = 39.22 gallons. U. S. 
~The palmo, long measure, = 9.725 inches, U. S. 
The braccio is 24 palmi == 22.692 inches, U. S. 
100 lb. peso grosso, VENICE, 105.18 lb avour’s. 
100 Jb. peso sottile = 66.4 lb. avoir’s. 
The stajo, a dry measure, = 2.27 bushels, U. S. 
The moggio is 4 staja —=9.08 bushels, U. S. 
The bigoncia, liquid measure, 34.2375 galls. Uo. 
The anfora is 4 bigonzi. = 136.95 galls. U. S. 
The braccio for woollens, = 26.61 inches, U. S. 
The braccio for silks — 24.8 inches, U. 8. 
The Venetian foot — 13.68 inches, U. S. 
100 lb. of TRIESTE, — 123.6 lb. avoirdupois. 
The stajo, dry measure, == 2.344 bushels, U. 8. 
The orna, or eimer, liquid = 14.94 gallons, U. S. 
The cil for woollens —926.6 inches, U.S. 
The ell for silks — 25.2 inches, U. S. 
The Austrian mile —=4.6 miles, U. S. 
100 lb. or libras, ROME, — 74.77 lb. avoirdupois. 
The rubbio, dry measure, = 8.356 bushels, U. S. 
The barile, liquid measure, = 15.409 galls. U. S. 
The Roman foot — 11.72 inches, U. S. 
The mercantile canna — 78.34 inches, U. S. 
The Roman mile —7.4 furlongs, U. S. 
100 Ib. or 100 rottoli, MALTA,=174.5 lb. avoirdupois. 
The salma, dry measure, —§.221 bushels, U. S. 
The foot of Malta —111 inches, U. 5. 
The canna is 8 palmi — 81.9 inches, U. S. 
The cantaro, kintal, SMYRNA,= 129.48 lb.avoirdupois. 
_ The oke or oka == 2.833 lb. avoirdupois. 
The killow, dry measure, — 1.456 bushels, U. 5. 


The pic, long measure, == 27 inches, U. S. 


256 ARITHMETIC. XXXVI. 
A factory maund of BENGAL, =742 lb. avoirdupois. 


A bazar maund, = 82; lb. avoirdupois. 
The haut or cubit == 18 inches, U. S. 
The guz ==1 yard U: S. 
The coss or mile = 1.238 miles U. S. 
The maund of BOMBAY, == 28 lb. avoirdupois. 
The candy is 20 maunds = 560 lb. avoirdupois. 
A bag of tice weighs 6 maunds —168 lb. avoirdupois. 
The candy, dry measure, == 25 bushels, U. S. 
The haut or covid — == 18 inches, U. S. 
The maund of MADRAS, = 25 lb. avoirdupois. 
The candy is 20 maunds = 500 lb. avoirdupois. 
The baruay, a Malabar weight, —482.25 Ib. avoir’s. 
The garee, dry measure, = 140 bushels, U. S. 
The covid, long measure, == 18 inches, U. S. 
The pecul of CANTON, = 1333 lb. avoirdupois. 
The catty is100th part of a pecul, = 1.333 lh. avoirdupois. 
The covid or cobre, long meas. == 14.623 inches U. 8. 
The pecul of JAPAN, = 130 lb. avoirdupois 
The catti is 100th part of a pecul, =1.3 ib. avoirdupois. 
The ine or tattamy, long meas. —6.25 feat,“Ue'S: 
The bahar of BENCOOLEN, =560 lb. avoirdupois. 
The bamboo, liquid measure, -==1 gallon, U. S. 

The coyang is 800 bamboo: == 800 gallons, U. S. 
‘The bahar of ACHEEN, = 423.425 lb. avoir’s. 
The maund of rice =75 lb. avoirdupois. 
The loxa of betel nuts = 10000 nuts. 

The loxa of nuts (when good) —168]b. avoirdupois. 
The pecul of BATAVIA, == 13572 lb. avoirdupois. 
33 kannes, liquid measure, 13 gallons, U. S. 
The ell, long measure, = 27 inches, U. S. 


The candy of COLOMBO, -=—=8500lb. avoirdupois. 


XX XIX MENSURATION. 257 


XXXIX. 
MENSURATION. 


MENSURATION is the art or practice of measuring, 
and has primary reference to the measurement of super- 
ficies and solids. 

Mensuration involves a knowledge of Geometry; and, 
as that science is not the object of this work, we shall 
confine our exercises under this head to,those measure- 
ments, which are most likely to be useful in the ordinary 
concerns of life. 


SUPERFICIES OR SURFACE. 


It has already been taught, that surfaces are measured 
in squares, and that the area of any square figure, or any 
parallelogram is found by multiplying together the length 
and breadth of the figure. For observations on the square 
and parallelogram, see page 162. 


Areaor A Ruomsus. <A: rhom- 
bus is a figure with four equal 
sides, having two of its angles 
greater, and two less than the 
angles of a square. The greater 
angles are called obtuse angles, 
and the smaller, acute angles. 
To find the area of arhombus, first drop a perpendicular 
from one of the obtuse angles to the opposite side, then 


“multiply the side by the perpendicular. 


1. How many square feet are there in a flooring, the 
form of which is that of a rhombus, measuring 15 feet on 
the side, and 12.5 feet in the perpendicular ? 


Area oF A Ruomsorp. A rhom- 
boid is a figure with four sides, 
which are not all equal, but . 
whose opposite sides are equal, 
and whose opposite eee te 


258 ARITHMETIC. XXXIX. 
equal, having, like a rhombus, two obtuse, and two acute — 
angles. ‘To find the area of a rhomboid, drop a perpen- 
dicular from one of the obtuse angles, to the opposite longer 
side, and multiply the longer side by the perpendicular. 

2. What is the area of a rhomboid whose longer side 
is 18.75 feet, and whose perpendicular is 9.25 feet ? 


Anes or TRIANGLES... Tt fcc Z 
is obvious, that a right-angled | : 
triangle contains just half as | 
much surface as would be con- | 
tained in a square or parallelo- | 
gram, two of whose sides are | 
formed by the base and perpen- | ———— 
‘dicular of the triangle. Therefore, the area of a right- 
angled triangle is found, by multiplying together either 
the base and half the perpendicular, or, the perpendicular 
and half the base. 

3. How many square rods of land are there ina lot, 
which is laid out in a right-angled triangle, the base mea- 
suring 19 rods, and the perpendicular 15 rods? 

4. Howmany acres of land in a lot, whose form is that 
of a right-angled triangle, the base measuring 113 rods, 
and the perpendicular 75 rods ? 


An Equilateral triangle is 
a triangle whose sides are all 
equal—such is the first of the 
two triangles adjoined. An 
obtuse-angled triangle is that 
which has one obtuse angle 
—~such is the second of the 
triangles adjoined. Whatever 
may be the form of a triangle, 
if it have not a right angle, it 
must be cut into two right- 
angled triangles before it can 
be measured: and this is done 
by dropping a perpendicular 
from the opposite angle to the 


XxXXIX. MENSURATION. 259 


base. The area is then found by multiplying together the 
base and half the perpendicular, or, the perpendicular 
and half the base. 

5. How many square inches in a triangle, whose base 
is 17} inches, and whose perpendicular height is 113 
inches ? 

6. How many square feet in a board 18 feet long, 16 
inches wide at one end, and tapering to a point at the 
other end? 

7. How many square feet in a plank 14 feet long, 17 
Inches wide at one end, and 10 inches wide at the other 
end? 

In this example, add the width of the two ends together, 
and take half the sum for one of the factors. 


Area or Circuies. To find the area of a circle, 
multiply the circumference by half the diameter, and 
divide the product by 2. When either the circumfer- 
ence, or the diameter is the only. dimension known, the 
other dimension may be found, as stated in page 173. 

8. What is the area of the head of a cask, the diameter 
of which is 18 inches ? 

9. Suppose a cylinder to measure 3 feet in circum- 
ference; what is the area of one end? 


Area oF Guozes. To find the convex area of a 
globe or sphere, multiply the circumference and diameter 
together. When the diameter is not known, it may be 
found from the circumference, as stated in page 173. 

10. How many square inches are there on the surface 
of a globe, whose circumference is 14 inches? 
_ 11. Suppose the earth to be 25020 miles in circum- 
ference, what must be the area of its whole surface ? 


SOLIDS AND CAPACITIES. 


It has already been taught, that solids and capacities are 
measured in cubes. It has also been shown, that the 
contents of any thing having six sides— its opposite sides 


260 ARITHMETIC. XXXKIX 


being equal, and all its angles being right angles— are 
found by multiplying together the length, and breadth, 
and depth of the thing. 


Souipiry of WepeceEs. To find the solid contents — 
of a wedge, first, find the area of the head or end of the 
wedge, and then multiply this area by half the length. 

12. How many solid inches are there in a wedge, 12 
inches long, 3 inches wide, and 14 inch thick at the head? 

13. What are the solid contents (in feet and inches) 
of a plank, 15 feet long, 17 inches wide, 234 inches thick 
at one end, and the thickness tapering to nothing at the 
other end? 

14. What are the solid contents of a stick of hewn 
timber, measuring in length 13 feet, in breadth 2ft. 4in., 
in depth 2 feet at one end, and 1 ft. 6in. at the other end ? 

In this example, add the depth of the two ends together, 
and take one half of the sum for the depth to be used — 
a the multiplication. . 


SoLipiTy or Prisms. A prism is a 
body with two equal ends, which are 
either square, triangular, or polygonal, 
and three or more sides, which meet in 
parallel lines, running from the several 
angles of one end-to those of the other. 
The adjoined is a representation of a 
triangular prism. 

The solid contents of prisms of all 
kinds, whether square, triangular, or 
polygonal, are found by one general rule, 
viz. Find the area of the end or base, and multiply this 
area by the length or height. 

15. How many cubic inches are there in a triangular 
prism, which is 16 inches in length, the ends measuring 
1.2 inches on a side, and 1.01 inches perpendicular ? 

16. How many cubic feet are there in a stick of tim- 
ber 18 feet long, hewn 3 square, the ends forming equt- 
lateral triangles of 10 inches side, and 8.7 inches perpen- 
dicular ? 


mx XIX. MENSURATION. 261 


Soxipity oF Cyiinpers. A cylin- 
der is around body, the two opposite 
sides, or ends of which, are circular 
planes, equal, and parallel. For instance, 
a stick of round timber of uniform cir- 
cumference, having its ends sawed at 
right angles with its length, is a cylinder: 
also, acommon grindstone is a cylinder. 
To find the solid contents of a cylinder, 
first, find the area of one end, and then 
multiply this area by the length. 

- 17. What are the solid contents of a cylinder whose 
length is 5 feet, and circumference 6.4 feet? (To find 
the diameter, see page 173.) c 

18. What are the contents of a cylinder whose length 
is 2 feet, and diameter 10 inches? 


SoLipity or Pyramips. Solids, which decrease 
gradually from the base, till they come to a point, are 
called pyramids. They are of different kinds, according 
to the figure of their bases. Ifthe pyramid has a square 
base, it is called a square pyramid; if a triangular base, a 
triangular pyramid; if the base bea circle, a circular 
pyramid, ora Cone. The point in which the pyramid 
ends is called the verter. A line through the centre of 
the pyramid, from the vertex to the base, is the height. 

The Frustrwm of a pyramid is what remains, after any 
portion of the top has been cut off, parallel to the base. 

f 


Square Pyramid. Triangular Pyramid 


262 ARITHMETIC. XXXIX. 


To find the cubical contents of a pyramid, -first find 


the area of the base, then multiply this area by one-third 
of the height. 
19. How many cubic inches are there in a square 


pyramid, 3 feet in height, and 9 inches square at the 


base ? 

20. How many cubic inches are there m a triangular 
pyramid, measuring 4 feet in height, 12 inches on each 
side of the base, and 10.4 inches from either angle of the 
base perpendicular to the opposite side? 

21. How many cubical inches in a cone, the height 


of which is 19 inches, and the diameter of the base 12 


inches ? 


SoLipity oF Frustrums. To find the cubical con- 


tents of the frustrum of a square pyramid, multiply the 
side of the base by the side of the top, and to the product 
add one-third of the square of the difference of the sides, 
and the sum will be the mean area between the two ends. 
Multiply the mean area by the height, and the product 
will be the cubical contents. 


To find the cubical contents of the frustrum of a Cone, 


multiply together the diameters of base and top, and to 
the product add one-third of the square of the difference 
of the diameters; then multiply this sum by .7854, and 
the product will be the mean area between the two ends. 
Multiply the mean area by the height, and the product 
will be the cubical contents. 

22. How many cubical inches in the frustrum of a 
square pyramid, 20 inches in height, 12 inches square at 
the base, and 5 inches square at the top? 

23. How many cubic feet in a stick of hewn timber, 
18 feet long, 16 mches square at one end, and 12} inches 
square at the other end? 


24. How many cubic inches are there in the frustrum — 


of a cone, measuring 3 feet in height, 16 inches in diam- 
eter at the base, and 6 inches in diameter at the top? 

25. How many gallons of water can be contained ina 
round cistern, 6 feet in height, 4 feet in diameter at the 
bottom, and 33 feet in diameter at the top? (Allow 231 
cubic inches to the gallon.) 


~XXXIx MENSURATION. 263 


Sotipity or Giozes. To find the cubical contents 
of a globe or sphere, first, find the convex area, as before 
directed, then multiply the area by one-sixth of the diam- 
eter; the product will be the cubical contents. 
26. What are the cubical contents of a globe measuring 
25 inches in circumference ? 
_ 27. How many cubic miles does the earth contam, 
allowing its circumference to be 25020 miles ? 


Sotipity oF IrrEeeuLAR Bopires. The cubical 
contents of a body, which cannot be reduced to regular 
geometrical form may be found as follows. Immerse tt 
in a vessel partly full of water; then the contents of that 
part of the vessel filled by the rising of the water will be 
the contents of the body vmmersed. 

28. How many cubic inches are there in a lobster, 
which, being immersed in a bucket 10 inches in diameter 
at top and bottom, raises the water 3 inches ? 


GAUGING OF CASKS. 


Although the difficulty of getting the true dimensions 
of the interior of casks, and the variety of their curve, 
must prevent perfect accuracy in their mensuration, yet, 
by careful observation in taking the dimensions, a result 
may be had, which will be sufficiently correct for all com- 
mon purposes. 

RULE. Take the interior length of the cask, the diame- 
ter at the bung, and the diameter at the head, all in 
inches. Subtract the head diameter from the bung 
diameter, and note the difference. 

If the staves of the cask be mucH curved between the 
bung and head, multiply the difference noted by .7; if 
but LITTLE curved, by .6; or, if they be of a MEDIUM 
curve, by .65; and add the product to the head diameter; 
the sum is the mean diameter, and thus the cask is re- 
_ duced to a cylinder. 

Square the mean diameter, and niultiply the square by 
the length of the cask; then divide this product by 294, 
and the quotient will be the number of wine gallons, which 
the cask may contain. 


264 ARITHMETIC. XXXIX 


It may be observed, that when a cask is reduced to a 
cylinder, its contents may be found ‘in cubical inches, 
and thence its contents in bushels, or any other of the 
measures of capacity. 

The length of the cask is most conveniently taken by 
ca.upers; allowing for the thickness of both heads, from 
1 to 2 inches, according to the size of the cask. When 
0 callipers can be had, the length of the stave must be 
taken in a right line, and a proper deduction made for 
the chimes, with that for the heads. The head diameter 
is to be taken within the chimes, and from .3 to .6 of an 
inch must be deducted, on account of the greater thick- 
ness of the stave inside the head. 

29. How many gallons will a cask contain, the interior 
of which measures 34.5 inches in length, 19 inches in 
diameter at the bung, and 16 inches in diameter at the 
head; the staves being much curved? 

30. How many gallons will a cask contain, the dimen- 
sions of which are 43 inches in length, 31.4 inches bung 
diameter, and 26 inches head diameter; the staves being 
but little curved ? 

31. Find the capacity of a cask measuring 52 inches 
in length, 33.5 inches bung diameter, 25.3 inches head 


diameter, and of medium curve between the bung ‘and 
head. : 


TONNAGE OF. VESSELS. 


There are two methods of measuring a vessel practised 
-——one by the ship-carpenter, who builds the vessel at a 
certain price per ton, and another by the officers of gov- 
ernment, who collect the revenue. 

CARPENTERS’ RULE. For single-decked vessels, mul- 
tiply together the length of the keel, the’breadth at the 
main beam, and the depth of the hold—all in feet—and 
divide the product by 95; the product isthe tonnage. For 
double-decked vessels, take half the breadth at the beam 
for the depth of the hold, and work as before. 

When a single-decked vessel has its deck bolted at any 
height above the wale, the carpenter is usually paid for 


a? 


~XXXIX. MENSURATION. 265 


. one-half of this extra height; that is, one-half of the height 

‘ above the wale is added to the depth below the wale, and 

this sum is used in the calculation, as the depth of the 
hold. 

GOVERNMENT RULE. ‘‘Jf the vessel be double-decked, 
take the length thereof from the fore part of the main 
stern, to the after part of the stern-post, above the upper 

deck; the breadth thereof at the broadest part above the 
main wales, half of which breadth shall be accounted the 
depth of such vessel, and then deduct. from the length, 

_ three-fifths of the breadth, multiply the remainder by the 
breadth and the product by the depth, and divide this 
last product by 95, the quotient whereof shall be deemed 
the true contents or tonnage of such ship or vessel; and if 
such ship or vessel be single-decked, take the length and 
breadth, as above directed, deduct from said length three- 
fifths of the breadth, and take the depth from the under 

- side of the deck plank to the ceiling in the hold, then 
multiply and divide as aforesaid, and the quotient shall 
be deemed the tonnage.”’ 


32. What is the carpenter’s tonnage of a single-decked 
vessel, the keel of which measures 60 feet, the breadth 
20 feet, and the depth 8 feet ¢ 

33. What is the carpenter’s tonnage of a double-decked 
vessel of 72 feet keel, and 22.5 feet breadth ? 

34. A merchant agreed with a carpenter to build a 
single-decked vessel of 58 feet keel, 20 feet breadth at 
the beam, and 8 feet hold, but afterwards chose to make 
the hold 10 feet deep, by raising the deck 2 feet above 
the wale. What tonnage must be paid for? 

35. What is the government tonnage of a double-decked 
vessel, 110.5 feet keel, and 30.6 feet breadth at the 
beam? . 

36. What is the government tonnage of a single-decked 
vessel, which measures 76.4 feet in length, 28.6 feet in 
breadth, and 12.3 feet in depth? é: 

37. What is the government tonnage of a single-decked 
vessel, whose length is 66 feet, breadth 20 feet, and depth 
9 feet? 

; 29 


266 ARITHMETIC. XL. 


ar XL. 
MECHANICAL POWERS. 


The MECHANICAL POWERS are certain simple instru- 
ments employed in raising greater weights, or overcoming 
greater resistance than could be effected by the direct 
application of natural strength. They are usually ac- 
counted six in number; viz. the Lever, the Wheel and 
/lele, the Pulley, the Inclined Plane, the Wedge, and, 
the Screw. 

The advantage gained by the use of the mechanical 
powers, does not consist in any increase of the quantum. 
of force exerted by the moving agent, but, in the concen- 
tration of force; that is, in bringing the whole force of a 
power acting through a greater space, into an action 
within a less space. The principle is illustrated by the 
consideration, that the quantum of force necessary to raise 
i pound 10 feet, will raise 10 pounds 1 foot. 

Weight and Power, when opposed to each other, sig- 
nify the body to be moved and the body that moves it. 


| 
) 
: 
| 


THE LEVER. 


A lever is any inflexible 
bar, which serves to raise , 
weights, while it is support- & 
ed at a point, which is the “™ 
centre of its motion, by a | 
julcrum or prop. There are several kinds of lever used 
in mechanics; the more common kind, however, is that 
which is shown above. : 

ls the distance between the weight and fulcrum is to 
the distance between the power and fulcrum, so is the 
power to the weight. 

It must he observed, that, in the above proportion, and 
in all the succeeding proportions of weight and power, 
the power intended is only sufficient to balance the weight. 
If the weight is to be raised, sufficient power must be 


XL. MECHANICAL POWERS. 267 


added to overcome friction; then any further addition of 
power will produce motion; and the comparative velocity 
of the weight and power, will depend on the comparative 
length of the two arms of the lever. It is a universal 
principle in mechanics, that the ratio of the power to the 
weight is equal to the ratio of the velocity of the weight 
io the velocity of the power . 

1. If a man weighing 160 pounds rest on the end of a 
lever 10 feet long, what weight will he balance on the 
other end, the fulcrum being 1 foot from the weight ? 

In this example, the distance between the weight and 
fulcrum being 1 foot, that between the power and fulcrum 
pwe10-—1— 9 ft. . Then 1ft. ; 9ft.=-160Ib:°: A 

2. Suppose a weight of 1440 pounds is to be raised 
with a lever 10 feet long, the fulcrum being fixed 1 foot 
from the weight; what power must be applied to the other 
end of the lever, to effect a balance ? 

(Oft. : 1ft.—1440]b. : A) 

3. If a weight of 1440 pounds be placed 1 foot from 
the fulcrum; at what distance from the fulcrum must a 
power of 160 pounds be placed, to balance the weight ? 

(1601b. : 1440]b.—=1ft. : A) 

4. At what distance from a weight of 1440 pounds must 
the fulcrum be placed, so that a power of 160 pounds, 
applied 9 feet from the fulcrum, will effect a balance ? 

| (1440]b. : 1601b.9ft. : A) 

5. If one arm of a lever be 44 feet, and the other 5 
feet, what power must be applied to the longer arm, to 
balance a weight of 500 pounds on the shorter arm Fare, 

6. Suppose a lever 6 feet long, with one end applied 
to arock, which weighs 1000 pounds, and resting on a 
fulerum 13 foot from the rock; what power must be ap- 
plied to the other end, to balance the rock? 

7. Suppose a bar 12 feet long to have 60 pounds at- 
tached to one end, and 30 pounds to the other, at what 
distance from each end must a fulcrum be placed, to 
produce a balance ? : 

g. If A and B carry a weight of 250 pounds, suspended 
upon a pole between them, 5 feet from A, and 3 feet 
from B, how many pounds does each carry ? 


268 , ARITHMETIC. XL. 


THE WHEEL AND AXLE. 


_ Tne wheel and axle are 
here represented, with the 
weight attached to the cir- 
cumference of the axle, and 
the power applied to the cit- 
cumference of the wheel. The 
principle of the lever is ob- 
vious In the wheel and axle— 
the axis or common centre 
being the fulcrum, the circum- 
ference of the wheel being the 
power end of the lever, and 
the circumference of the axle, 
the end applied to the weight. Hence, the radius of the - 
axle is to the radius of the wheel, as the power is to the - 
weight: or, by a statement more frequently convenient— _ 

/Is the diameter of the azle is to the diameter of the 
wheel, so is the power to the weight. | 

9. Amechanic would make a windlass in such manner, 
that 1 pound applied to the wheel, shall he equal to 10 
pounds suspended from the axle. Now, supposing the 
axle to be six inches in diameter, what must be the diam- 
eter of the wheel ? 

10. Suppose the diameter of a wheel to be 8 feet, what 
must be the diameter of the axle, that 1 pound on the 
wheel shall balance 15 pounds on the axle ? 

11. Suppose the diameter of an axle to be 4 inches, 
and that of the wheel 3 feet; what power at the wheel 
will balance 28 pounds at the axle? 

12. If the diameter of a wheel be 7 feet, and that of 
the axle 8 inches, what weight at the axle will balance 
40 pounds at the wheel ? pay | 

13. There are two wheels; one of which is 6 feet in 
diameter, with an axle of 9 inches diameter; and the other- 
is 4 feet in diameter, with an axle of 7 inches diameter. 
Suppose the power cord of the smaller wheel to be coiled 
upon the axle of the larger; what weight on the axle of — 
the smaller wheel would be balanced 100 at the power 
cord of the larger wheel ? 


oe 


single or combined: it is also, 
_either fixed, or movable. If 


XL. MECHANICAL POWERS. 269 


THE PULLEY. 


A pulley is a small wheel, 77a TIT 
which turns on an axis pass- 


ing through its centre and 
fixed in a block, receiving 
its motion from a cord, that 
passes round its circumfer- 
ence. The pulley is either 


a power sustain a weight by 
means of a single, fixed pul- 
ley—a cord passing over it, 
with the weight attached to 
one end and the power to the 
other—the power and weight 
are equal: and if the pulley 
be put in motion, the velocity a 
of the power, and theveloci- “" ~~ 58 
ty of the weight will also be equal. But, if the fixed 
pulley be combined with one movable pulley—as repre- 
sented in the first set of pulleys above—the weight is 
equal to twice the power which sustains it; and if the 
pulley be put in motion, thevelocity of the power will be 
equal to twice the velocity of the weight. Thus, every 
cord going over a movable pulley, adds 2 to the powers, 
and hence, in a system of pulleys, we have the following 
proportion. 

As 1 is to twice the number of movable pulleys, so is 
the power to the weight. | 

14. In the second set of pulleys represented above, 
three of the pulleys are fixed, and three are movable. If 
a power of 45 pounds were applied to the cord, what 


_ weight would it balance? 


15. What power must be applied to a cord that runs 
over 2 movable pulleys, in order to balance a weight of 
800 pounds ? 

16. What power must be applied to a cord that runs 
over 6 movable pulleys, to balance a weight of 2000 
pounds ° : -23* 


270 ARITHMETIC. DO ii 


| 
| 
| 


17. Ifa cord, which runs over 3 movable pulleys, be — 


attached to an axle 4 inches in diameter, the wheel of the 
axle being 38 inches in diameter, and a power of 20 
pounds be exerted at the circumference of the wheel, 
what weight would be raised under the pulleys? 


THE INCLINED PLANE. 


An inclined plane is a plane 
making an angle with the hori- 
zon. For instance, a plank 
presents an inclined plane, 
when one end is resting upon 
coe 
end is raised to any height less <i 
than that which would render it vertical. A convenient 
use of the inclined plane is exemplified in rolling casks 
from a cellar, upon sloping pieces of timber, or planks. 

On an inclined plane, as the perpendicular height of 
the plane is to the length of the plane, so is the power to 
the weight.. 

18. A certain inclined plane is 16 feuet in length, and 7 
feet in perpendicular height. What weight might be 
drawn up this plane, by a power, which, if exerted ona 
cord over a single, frxed pulley, would raise 25 pounds ? 

19. What power would be necessary to sustain a roll- 
ing weight of 1000 pounds, upon an inclined plane of 75 
feet length, and 38 feet perpendicular height ? 


20. What must be the length of an inclied plane, whose — 


perpendicular height is 15 feet, that the exertion of the 
power of 42 pounds shall draw up 200 pounds ? 

21. On a rail-road, there is an inclined plane of 80 
rods in length, rising to a perpendicular height of 50 feet. 


What power must be exerted on the summit, to draw up — 


a train of cars weighing 62000 pounds ? 


22. Suppose a set of pulleys, 3 of which are movable, — 


to be applied to a weight upon an inclined plane of 50 
feet length, and 14 feet perpendicular height; what weight 


upon the plane, would be sustained by 40 pounds at the — 


power cord of the pulleys ? 


as amoving inclined plane; the 


of the plane. In the wedge, 


XL. MECHANICAL POWERS. ~ Bat 


THE WEDGE. 
The wedge may be viewed 


Ee 
= 
f 


head of the wedge, where the 
power is applied, answering 
to the perpendicular height 


however, the inclined plane is | HI : 
double, and the force produce 
double, andthe force prod 
two equal parts, acting at right angles with each side. 

As the breadth of the head of a wedge is to the length 
of its side, so is the power acting against the head, to the 
force produced at the side. 

Observe, that the force mentioned in the above pro- 
portion, respects one side of the wedge, only. If the 


_ forces against both sides be required, then, only half the 


breadth of the head must be taken into the proportion. 
In the common mode of applying the wedge, the fric- 
tion against the sides is very great— at least equal to the 
force to be overcome. Therefore, not less than one-half 
of the power is lost; and for this loss there is no allow- 
ance made in the above proportion The wedge, how- 
ever, has a great advantage over all the other mechanical 
powers, arising from the force of percussion or blow with 
which the head is struck, by a mallet. ‘The power thus 


- obtained is incomparably greater than that of any dead 


weight or pressure, such as is commonly employed on 

other instruments. 
23. Suppose a power of 50 pounds to be applied to a 

wedge, the head of which is 2 inches broad, and the side 


12 inches long, what weight of force would be effected — 


on either side; if there were no friction to resist ? 
24. Ifa force of 1000 pounds is to be effected on the 
side of a wedge, that is 14 inches long, and 3 inches 


broad at the head, what power must be applied to the 
head; allowing nothing for friction? Again, allowing the 
friction, which is to be overcome, to be equal to the 


force effected, what power will be necessary ? 


272 ARITHMETIC. XL. 


THE SCREW. 


The screwis a spiral thread 
or groove, cut-round a cylin- 
der, and every where mak- 
ing the same angle with the 
length of the cylinder. In 
oneround ofthe spiral,itrises pee u 
along the cylinder, the dis- }EH 
tance between two threads. 
Therefore, if the surface of 
the cylinder, with the spiral 
thread on it, were unfolded 
and stretched into a plane, 
the spiral would form a 
straight inclined plane, whose tenet Sot be to its 
height, as the circumference of the cylinder is to the dis- 
tance between two threads of the screw. The inclined 
plane being thus recognised in the screw, the following 
proportion is obvious. 

As the distance between two threads of a screw is to 
the circumference of the circle described by one revolu- 
tion of the power, so is the power to the weight. 

The length of the lever to which the power is applied, 


eI 
ii 
Hi! 
1H | 
| 
==! 
iHiszs! 

Sal 


being one- -half of the diameter of the circle round which 


the power revolves, the circumference may be found from 
the lever, as taught in page 173. 

In the common use of the screw, about one-third of 
the power is expended in overcoming friction; and for 
this loss, no allowance is made in the above stated pro- 
portion. 

25. If the threads of a screw be 1 inch apart, and a 
power of 50 pounds be exerted at the end of a lever 70 
inches long, what weight of force will be produced at the 
end of the screw; allowmg nothing for friction. — 

26. If the threads of a screw be .2 of an inch apart, 


and a power of 40 pounds be exerted at the end of a. 
lever 30 inches long, what will be the force at the end — 
of the screw; allowing 4 of the power to be lost in over- 


coming friction ? 


XLI. MISCELLANEOUS QUESTIONS. are 


27. Suppose a power of 48 pounds is to be employed 
to effect the weight of 5000 pounds, by means of a screw, 
whose threads are 1.3 inches apart; what must be the 


length of the lever; allowing 4 of the power to be lost in 


overcoming friction ? 

28. Suppose the end of a screw, whose threads are 
.8 of an inch apart, and whose lever is 7ft. long, to be set 
upon a wedge, that is 15in. long at the side, and 2 inches 
broad at the head; what weight of force would be effected 
on either side of the wedge, by applying 100 pounds’ 


power to the lever; allowing 4 of the force on the screw, 


and 3 of that on the wedge to be lost in friction ? 


~» XDI. 


MISCELLANEOUS QUESTIONS. 


1. What vulgar fraction is that, which being multiplied 
by 15, will produce 3? 

2. What decimal fraction is that, which being multi- 
plied by 15, will produce .75? 

3. What quantity is that, which being divided by 3, 
gives the quotient 21? 

4. What vulgar fraction 1 is that, from which if you take 


3, the remainder will be }! 


5.. What vulgar fraction ‘s that, to which if you add 2, 
the sum will be 2° 

6. What quantity is that, which bemg ee by 2, 
produces the fraction }? 

7. What quantity is ; that, from which if you take 2 of 
itself, the remainder will be 12? 

8. What quantity is that, to which if you add 2 of ; 
of itself, the sum will be 61? 

9. A farmer carried to market a load of produce, con- 


) sisting of 780]b. of pork, 250lb. of cheese, and 154 |b. 
of butter; he sold the pork at 6 cents, the cheese at 8 
cents, and the butter at 15 cents per Ib.; ; and agreed to 


take in pay, 601b. of sugar at 10-cents per lb., 15 gallons 


O74 ARITHMETIC. Kael: 


of molasses at 40 cents a gallon, 4 barrel of mackerel at 
$3.50, 4 bushels of salt at 90 cents a bushel, and the 
balance in cash. How much money did he receive? _ 


10. A and B commenced business with equal sums of © 


money; A gained a sum equal to + of his stock, but B lost 
$200, and then had only half as much as A. What was 
the original stock of each ? 


11. A’man was hired for a term of 50 days on condi- 


tions, that for every day he worked he should receive 
75 cents, and for every day he was idle he should pay 25 
cents for his board; at the expiration of the time, he was 
entitled to $27.50. How many days was he idle ? 


12. A and B have the same income; A saves 4 of his; — 


but B, by spending $30. a year more than A, at the end 


of 8 years finds himself $40. in debt. What is their — 


income, and what does each spend a year? 

13. A grocer has two sorts of tea; one at 75 cents ia 
pound, and’the other at $1.10 a pound. In what pro- 
portion must he mix them, in order to afford the mixture 
at $1. a pound? 


14. A and B can doapiece of work in 5 days; A alone — 


can do it in 7 days. In what time can B do it? 

15. After A has travelled 51 miles, B sets out to over- 
take him, and travels 19 miles to A’s 16. How many 
miles will each have travelled, before B overtakes A? 


16. A trader bought a cask of wine, but, in conveying ~ 
it home, 4 of it leaked out. He sold the remainder, at — 
$2.50 a gallon, and thus received what he paid for. the — 


whole. How much per gallon did he give for it ? 

17. A person having spent in one year all his mcome 
and + as much more, found that by saving 3/5. of his in- 
come afterward, he could, in 4 years make good the de- 
ficiency, and have $20 left. What was his income? 


18. A young hare starts 40 yards before a grey-hound, 


and is not perceived by him till she has been up 40 sec- 
onds; she scuds away at the rate of 10 miles an hour, 
and the hound, in view, makes after her at the rate of 18 
miles an hour. How long will the course continue, and 
what will be the length of it, from the place where the 
hound set out? 


: 


“ 


. 


XLI. MISCELLANEOUS QUESTIONS. Q275 


19. A man driving his geese to market, was met by 
another, who said— ‘Good morning, with your hundred 
geese.’ * He replied— ‘I have not a hundred; but if I 
’ had half as many more than I have, and two geese and a 
half, I should have a hundred.’ How many had he? 

20. If 8 men can build a wall 15 rods long in 10 days, 
how many men will it take to build a wall 45 rods long in 
5 days? 

21. A gentleman had £7 17s. 6d. to pay among his 
laborers; to every boy he paid 6 pence, to every woman 
8 pence, and to every man 16 pence; there was one boy 
to three women, and one woman to two men. What 
was the number of each? 

22. A farmer bought a yoke of oxen, a cow, and a 
sheep for $82.50; he gave for the cow 8 times as much 
as for the sheep, and for the oxen 3 times as much as for 
the cow. How much did he give for each? 

23. The head of a fish was 9 inches long, its tail was 
as long as its head and half its body, and its body was as 
long as its head and tail both. What was the whole 
Jength of the fish? : 

24. The remainder of a division is 325, the quotient 
467, and the divisor is 43 more than the sum of pee 
what is the dividend ? 

25. A trader bought a hogshead containing 120 wallohe 
of molasses for 42 dollars. At what price per g gallon must 
he sell it, to gain 15 per cent.? 

26. Sold goods to the amount of $3120, to be paid 

one half in 3 months, and the other half in 6 months. 
- How much must be discounted for present payment, when 
-money is worth 6 per cent. a year? 
_ 27° A merchant imported 10 tons of iron at 95 dollars 
per ton; the freight and duties amounted to 145 dollars, 
and other charges to 25 dollars. At what price per |b. 
must he sell, to gain 20 per cent.? 

28. The hour and minute hands of a watch are together 
at 12 o’clock; when are they next together ? 

29. Suppose two steamboats to start at the same. time 
from places 300 miles apart on the same river; the one 
proceeding up stream is retarded by the current 2 miles - 


276 ARITHMETIC. XLI. 


yer hour; the other moving down stream is accelerated 
the same. If each is propelled by a steam engine, that 
would move it 8 miles an hour in still water, how far from 
each starting place will the boats meet? 

20. Thomas sold 150 pineapples at 335 cents apiece, 
and took no more money than Harry did for watermelons 
at 25 cents apiece. How much money did each take, 
and how many melons had Harry ? 

31. Seven-eighths of a certain number exceeds four- 
fifths of the same number by 6. What is the number ? 

32. If 18 grains of silver will make a thimble, and 12 
dwt. a teaspoon, how many thimbles and. teaspoons, of 
each an equal number, can be made from 150z. 6dwt.? 

33. What are the superficial contents of a piece of 
wainscot 8 ft. 64in. long, and 2ft. 97in. broad? 

34. A guardian paid his ward $3500. for $2500. 

. which he had in his hands 8 years. What rate of interest 
did he allow? 

35. A set out from Boston for Hartford precisely at 
the time, when B at Hartford set out for Boston, distant 
‘100 miles: after 7 hours they met on the road, and it 
then appeared, that A had ridden 13 mile an hour more 
than B. At what rate an hour did each travel ? 

36. A father divided his fortune among his sons, giving 
A $4 as often as B 3, and C 5 as often as B 6. What 
was the fortune, supposing A’s share to be $5000. ? 

37. A prize of 945 dollars is to be divided among a 
captain, 4 men, and a boy; the captain is to have a share 
and ahalf; the men each a share; and the boy 4 of a share. 
What ought each person to have ? ! 

38. A person left 40 shillings to four poor widows;_ 
viz. to A he left 4, to B 4, to C 4, and to D , desiring 
the whole might be distributed accordmgly. What is the 
proper share of each? 

39. A person looking on his watch, was asked what 
was the time of day; he answered—lIt is between 4 and_ 
5, and the hour and minute hands are exactly together. 
What was the time ? 

40. Divide 1200 acres of land among A, B, and C, 
so that B may have 100 acres more than A, and C 64 
acres more than B. - grits # 


X LI. MISCELLANEOUS QUESTIONS. SU 


41, What length of a board, which is 83 inches wide, 
will contain as much as a square foot ? 

42. What number is that, from which if you take 4 of 
3, and to the remainder add {3 of 3g, the sum is 10? | 

43. A can do a piece of work alone in 10 days, and 
B in 13 days. If both set about it together, in what time 
will it be finished ? 

44. A,B, and C were to share $100000. in the propor- 
tion of 4, 4, and +4, respectively; but C’s part being lost 
by his death, it is required to divide the whole sum prop- 
erly between the other two. 

45. Inan orchard of fruit trees, } of them bear apples, 
1 pears, 4 plums, and 50 of them cherries. How many 
trees are there in the orchard ? 

46. A cistern, containing 60 gallons of water, has 3 
unequal faucets for discharging it; the greatest faucet will 
empty it in one hour, the second in two hours, and the 
third in three hours. In what time will it be emptied, 
if they all run together ? 

47. What sum of money will amount to 336 dollars 
_ 42 cents in a year and 4 months, at 6 per cent. per an- 
num, simple interest ? 

48. A man, when he married, was 3 times as old as 
his wife; 15 years afterward he was but twice as old as 
his wife. At what age was each married ? | 

49. Divide 1000 dollars among A, B, and C, so as to 
give A 120 dollars more, and B 95 dollars less, than C. 

50. What fraction is that, to which if # of 3 be added, 
the sum-will be 1° 

51. A certain cubical stone contains 389017 solid feet. 
What are the superficial contents of one side ? 

52. A father dying left his son a fortune, 4 of which 
he spent in 8 months; 3 of the remainder lasted him 12 
months longer; after which he had only 1200 dollars left. 
How much did his father leave to him? 

53. Three travellers met at an inn, and two of them 
brought their provisions along with them; but the third 
not having provided any, proposed to the other two, that 
they should all eat together, and he would pay them for his 
proportion. This being agreed to, A produced 5 loaves, 

24. 


278 ARITHMETIC. XLL 


and B 3 loaves, which the travellers ate together, and C 
paid 8 equal pieces of money as the value of his share, 
with which the other two were satisfied, but quarreled 
about the division of them. Upon this, the affair was 
referred to an umpire, who decided the dispute justly. 
What was his decision ? 

54. What number is that, which being added to {4 of 
765, the sum will be equal to the square root of 2601? 

55. ‘wo persons talking of their ages, one says, 2 of 
my age is equal to ¢ of yours, and the difference of our | 
ages is 10 years. What were their ages? 

56. A man bought some lemons at 2 cents each, and 
¢ as many at 3:cents each, and then sold them all at the 
rate of 5 cents for 2, and thus gained 25 cents. How 
many lemons did he buy? 

57. There are two cisterns, which are constantly re- 
ceiving an equal quantity of water; but the first constantly 
loses ¢ of what it receives. After running 7 days, 10 bar- 
rels were taken from the second, and then the quantity of 
water in the two was equal. How much water did each 
receive per day? 

58. A person being asked the hour of the day, said, 
the time past noon is equal to 4 of the time to midnight. 
What. o’clock was it? 

59. What number, added to J; of 3813, will make the 
sum 200? 

60. A general forming his array into a square, finds he 
nas 284 soldiers over and above a square; but increasing 
each side with one soldier, he wants 25 to fill up a square. 
How many soldiers had he? 

61. A reservoir for water has two pipes to supply it; 
by the first alone it may be filled in 40 minutes, by the 
second alone in 50 minutes; and it has a discharging pipe, 
by which it may, when full, be emptied in 25 minutes. 
Now, if these three pipes were all left open, the influx 
and efflux of the water being always at the aforesaid rates, 
in what time would the cistern be filled ? 

62. Three persons do a piece of work; the first and 
second together do 7 of it, and the second and third 
together do {7 of it. What part of it is done by the 
second ? 


XLI. MISCELLANEOUS QUESTIONS. | 279 


63. A man driving some oxen, some cows, and some 
sheep, being asked how many he had of each sort, an- 
swered, that he had twice as many sheep as cows, and 
three times as many cows as oxen; and that the whole 
number was 80. What was the number of each sort ? 

64. A man has a note of $647. due in 2 years and 7 
months without interest; but being in want of money, he 
will sell the note; what ought he to receive, when interest 
is 6 per cent. a year? 

65. A gentleman bequeathed an estate of $12500. to 
his wife and son. The son’s share was % of the wife’s 
share. What was the share of each? 

66. A man and his wife found that when they were 
together, a bushel of corn would last them 15 days; but 
when the man was absent, it would last the woman alone 
27 days. How long would it last the man alone ¢ 

67. A farmer sold some calves and some sheep for 
$108.; the calves at $5. and the sheep at $8. apiece. 
There were twice as many calves as sheep. What was 
the number of each sort ? 

68. A owes B $158.33 due in 11 months and 17 days, 
without interest, which he proposes to pay at present. 
What ought he to pay, money being 5 per cent.? 

69. At what time, between twelve and one o’clock, do 
the hour and minute hands of a clock or. watch point in 
directions exactly opposite ° 

70. If 3 men can do a piece of work in 56 days, and 
4 women can do the same in the same time, in what time 
will one man and one woman together perform it? 

71. A son having asked his father’s age, the father 
thus replied; ‘your age is 12 years, to which if five- 
~ eighths of both our ages be added, the sum will express 
my age.’ What was the father’s age ? | 

72. Three gentlemen agree to contribute $730. to- 
wards the building of a church at the distance of 2 miles 
from the first, 22 miles from the second, and 33 miles 
from the third; and they agree, that their shares shall be 
reciprocally proportional to their distances from the 
church. How much must each contribute ? 

73. If Acanreap a field in 13 days, and B in 16 days, 
in what time can both together reap it ? | 


280 s ARITHMETIC. XLI. 


74. A and B set out together from the same place, and 
travelled in the same direction. A travelled uniformly 
18 miles a day, but after 9 days turned and went back as 
far as B had travelled during those 9 days; he then turned 
again, and, pursuing his journey, overtook B in 221 -days 
from the time they first set out. At what rate per day 
did B uniformly travel ? ; 

75. Two men, A and B, are on a straight road, on 
the opposite sides of a gate; A is distant from it 308 


yards, and B 277 yards, travelling each towards the gate. — 


How long must they walk, to make their distances from 
the gate equal; allowing A to walk 21 yards, and B 2 
vards, per second? ’ 

76. IT want just an acre of land cut off from the end of 
a piece, which is 135 rods wide; how much of the length 
of the piece will it take ? 


77. A farmer had oats at 38 cents a bushel, which he - 


mixed with corn at 75 cents a bushel, so that the mixture 
might be 50 cents a bushel. What were the proportions 
of the mixture ? 


78. A grocer mixed 123]b. of sugar worth 8 cents . 


per lb. with 871b. worth 11 cents per lb. and 15 lb. worth 
13 cents per lb. What was the mixture worth per lb.? 
79. A man travelling from Boston to Philadelphia, a 


distance of 335 miles, at the expiration of 7 days found. 


that the distance which he had to travel was equal to 23 
of the distance, which he had already travelled. How 
many miles per day did he travel ? : 

80. A gentleman bequeathed an estate of $50000. to 
his wife, son, and daughter; to his wife he gave $1500. 
more than to the son, and to his son $3500. more than 
to his daughter. How much was the share of each? 

81. The stock of a cotton manufactory is divided into 
32 shares, and owned equally by 8 persons, A, B, C, 
&c. A sells 3 of his shares to a ninth person, who thus 
becomes a member of the company, and B sells 2 of his 
shares to the company, who pay for them from the common 
stock. After this, what proportion of the whole stock 
does A own ?” 

82. How many feet in a stock of 18 boards, 12 feet 
3 inches long, and 1 foot 8 inches wide? 


& 


XLI. MISCELLANEOUS QUESTIONS. 281 


83. A merchant laid out $50. for lmen and cotton 
cloth, buying 3 yards of linen for a dollar, and 5 yards 
of cotton for a dollar. He afterwards sold of his linen 
and + of his cotton for$12, which was 60 cents more 
than it cost him. How much of each did he buy ? 

84. If 157 dollars 50 cents in 16 months gain 12 dol- 
lars 60 cents, in what time will 293 dollars 75 cents gain 
11 dollars 75 cents, at the same rate of interest? 

85. A merchant having goat-skins, and wishing to get 
some of them dressed, delivered for that purpose 560 to 
a currier, to be dressed at 124 cents each, who agreed to 
take his pay in dressed skins at 50 cents each. How 
many dressed skins should the currier return ? 

86. If eggs be bought at the rate of 5 for 4 cents, 
how must they be sold per dozen, to gain 25 per cent. ? 

87. What is the circumference of a wheel, the diameter 
of which is 5 feet ? : 

88. A lion of bronze, placed upon the basin of a foun- 
tain, can spout water into the basin through his throat, 
his eyes, and his right foot. If he spouts through his 
throat only, he will fill the basin in 6 hours; if through 
his right eye only, he will fill it in 2 days; if through his 
left eye only, in 3 days; if through his foot only, he will 
fill it in 4 hours. In what time will the basin be filled, 
if the water flow through all the apertures at once? 

89. A man having 100 dollars spent part of it, and 
afterward received five times as much as he had spent, 
‘and then his money was double what it was at first. How 
much did he spend? 

90. A hare starts 50 leaps before a grey-hound, and 
takes 4 leaps to the hound’s 3; but 2 of the hound’s leaps 
are equal to 3 of the hare’s. How many leaps must the 
hound make, to overtake the hare ? 

91. A grocer would mix the following kinds of sugar, 
viz. at 10 cents, 13 cents, and 16 cents per lb. What 
quantity of each must he take, to make a mixture worth 
12 cents per |b.? | 

92. A grocer has 43 gallons of wine worth $1.75 a 
gallon, which he wishes to mix with another kind worth 
$1.40 a gallon, in such proportion that the mixture may 

24* | 


982 ARITHMETIC. ate Sa, 


be worth $1.60 a gallon. How many gallons at $1.40 
must he use? 

93. Three merchants, A, B, and C, freight a ship with 
wine. A puts on board 500 tons, B 340 tons, and C 
94 tons; and in a storm they are obliged to cast 150 tons 
overboard. What loss does each sustain? | 

94. A and B hired a pasture for 37 dollars. A put 
in 3 horses for 4 months, and B 5 horses for 3 months. 
What ought each to pay? 

95. A family of 10 persons took a large house for 4 
of a year, for which. they agreed to pay 500 dollars for 
that time. At the end of 14 weeks, they took in 4 new * 
lodgers; and after 3 weeks, 4 more; and so on at the end 
of every 3 weeks, during the term, they took’in 4 more. 
How much rent must one of each class pay ? 

96. A boy bought 12 apples and 6 pears for 17 cents, 
and then, at the same rate, 3 apples and 12 pears for 20 
cents. What was the price of an apple, and of a pear? 

97. A certain square pavement contains 48841 square 
stones, all of the same size. How many stones constitute 
_ the length of one side of the pavement ? : q 

98. A certain field liesgin the form of a right-angled — 
triangle; the sides containing the right angle are, one 48 
rods, the other 20 rods in length. What is the length of 
_the other side? How many acres in the field ? 

99. The following lots of sugar, from Havana, were 
sold in Boston on account of owners in Cuba, at 123 
cents per lb. Required the amount of sales for each’ 
owner, allowing draft 4b. per box, and tare 15 per cent. 

A’s sugar, 21 boxes, weighing 10794 |b. gross. 
- B’s sugar, 70 boxes, weighing 35980]b. gross. 
C’s sugar, 84 boxes, weighing 43176 lb. gross. 
D’s sugar, 105 boxes, weighing 53970|b. gross. 

100. How much money on interest at the rate of 6 
per cent..a year, from February 16th 1835, will be suffi- 
cient to meet a custom-house bond of $1464.45, which 
will become due on 10th of January, 1836? 

101. How many shingles will cover the roof of a house, 
which is 40 feet in length, and has 30 feet rafters, sup- 
posing each shingle to be 4 inches wide, and each course 
to be 6 inches ? 


XLI. MISCELLANEOUS QUESTIONS. 283 
102. A merchant sold a piece of cloth for $40, and 


‘by so doing, lost 10 per cent. For how much should he 


have sold it, to have gained 15 per cent.? 


% 


103. A merchant received on consignment, three par- 
cels of hops, viz. 4501b. from Allen, 8901b. from Brooks, 
and 510lb. from Chase. Allen’s hops were found on 
inspection to be 334 per cent. better than the others, but 
it was necessary to sell them together, at 12 cts. a pound. 
How much must each owner be credited? ; 

104. Three parcels of beef, of 60 barrels each, were 
received at Baltimore, from Boston, marked, W, X, Y. 
The lot marked W was found to be 50 per cent. better 
than the others. The whole was sold together at 10 
dollars a barrel. How must the sale be adjusted between 
the owners of the beef ? a 

105. If iron worth $4, per cwt. cash, is sold for $4.50, 
on a credit of 8 months, what credit should be allowed 
on wine worth in cash $224 per pipe, but sold at $242, 
to make the percentage equal to that on the iron? 

106. The number of terms in an equidifferent series is 
11, the last term is 32, and the sum of the terms is 187. 
Find the first term, and the common difference. 

107. Amerchant has three notes, due to him as follows; 
one of $300, due in 2 months; one of $250, due in 5 
months; and one of $180, due 3 months ago, with inter- 
est; the whole of which he now receives. What sum is 
received on the three notes, allowing money to be worth 


6 per cent. a year? 


108. A lady has two silver cups, and only one cover. 
The first cup weighs 120z. _ If the first cup be covered, 
it will weigh twice as much as the second; but if the 
second cup be covered, it will weigh three times as mucli 
as the first. What is the weight of the cover, and of the 
second cup? 

109. Gray of Baltimore remits to Degrand in Boston, 
for sale, a set of exchange on London, the proceeds of 
which to be invested in certain merchandise for Gray’s. 
account. On selling the bill at 10 per cent. advance, 
D received $8600. How many pounds sterling was the 
bill drawn for, and how much is D to lay out for G, re- 


284 ARITHMETIC. XLL¥ 


‘ 


serving to himself 5 of 1 per cent. on the sale of the bill, 
and 24 per cent. commission on the investment ? 

110. The greatest term ina series of continual propor- 
tionals is 10, the ratio 14, and the number of terms 12. — 
What is the sum of the series? 

111. What is the area of a circle, the diameter of 
which is 200 feet ? 7@ 

112. What sum of money must be put on interest, at 
the rate of 6 per cent. a year, to gain $27.83 in 112 
months ? } 

113. A person found two sums of money; 4 of the 
first added to 4 of the second was $120. The two sums 
together were $400. What was each sum? 

114. What number is that, whose cube root is equal 
to the square root of 361? 

115. If a family of 9 persons spend $305. in 4 months, 
how many dollars would maintain them 8 months, if 5 
persons more were added to the family? 

116. Bought 5hhds. of wine at 1 dollar per gallon, 
cash; having kept it 3 months and 23 days, I sold it at 
$1.20 per gallon, on a credit of 5 months; 16 gallons 
having leaked out while in my possession. What was 
my cash gain? 

117. A grocer having sugars at $12, $10, and $8 per 
ewt. would make a mixture of 30 cwt. worth $9 per cwt. 
What quantity of each must he take ? 

118. What sum of money on interest at 6 per cent. a 
year, will amount to $1295.19, m 13 months 6 days? 

119. How much must be paid for the transportation 
of 72613lb. 60 miles, at the rate of $10 for the trans- 
portation of 2000lb. 47 miles ? 

120. A farmer sold 17 bushels of rye, and 13 bushels 
of wheat for $31.55. The wheat, at 35 cents a bushel 
more than the rye. What was each per bushel ? 

121. A man bought apples at 5 cents a dozen, half of 
which he exchanged for pears, at the rate of 8 apples for 
5 pears; he then sold all his apples and pears at a cent 
apiece, and thus gained 19 cents. How many apples 
did he buy, and how much did they cost ? | 

122. The sides of two square pieces of ground are as 


XLI - MISCELLANEOUS QUESTIONS. 285 


3 to 5, and the sum of their superficial contents is 830600 
square feet. What is the length of a side of each piece ? 

123. If 96 boards, 15 feet 6 inches long and 14 inches 
wide, will floor a place, how many will it take if the boards 
are only 11 feet 4 inches long and 9 inches wide? 

124. A certain club spent at a supper, 43 dollars 56 
cents, and the expense.of each was as many cents as there 
were persons in the company. What did each pay? 

125. If 20 feet of iron railing weigh 1000 1b. when the 
bars are 14 inch square, what will 50 feet come to, at 94 
cents per lb. when the bars are % of an inch square ? 

126. A stationer sold quills at $1.83 per thousand, 
and gained 4 of the first cost; but quills growing scarce, 
he raised the price to $2.04 per thousand. What did 
he gain per cent. by the last sale ? 

127. A merchant purchased goods to the amount of 
3472 dollars, which he sold at a loss of 124 per cent. 
and invested the proceeds of the sale in other goods, which 
he sold at a profit of 13 per cent. Did he gain or lose 
by these transactions, and how much? 

128. A house completely finished, has cost the owner 
$12894; it is 4 stories high, and the ground floor is 
divided into two shops, one of which is let at $225, the 
other at $200 a year; the three upper stories are let for 
$450 a year; the annual expense for repairs is $36.89. 
What per cent. does the house pay? 

) 129. Amerchant in Boston received from New Orleans 

a bill at 30 days sight; he allowed 1 per cent. discount 
_ for present payment, and received $2530.44. What sum 
was the bill drawn for; and what was the discount? 

130. A merchant sold a parcel of coffee at 15 cents 
per Ib. and lost 10 per cent.; soon after he sold another 
parcel, to the amount of $525. and gained 40 per cent. 
How many pounds were there in the last parcel; and at 
what price per lb. was it sold? . 

131. G received from H 760]b. of rough tallow to try 
out, at 60 cents per 1001b. clear, and was to take his pay 
in rough tallow at 8 cents per Ib.; G returned 615]b. clear, 
and H paid him the balance due to him in rough tallow. 
Allowing 18 per cent. for waste, what was the balance 
due to G? , 


coal 


w 


986 | ARITHMETIC. XL 


-132. A merchant received on consignment 3 lots of — 
hops, viz. 810]b. from Allen; 720]b. from Bond; and_ 


1872lb. from Cook. On inspection, Bond’s hops were 
found to be 123 per cent. better than Allen’s; and Cook’s 


5 per cent. better than Bond’s. A sale of the 3 lots to- — 


gether was effected at 10 cents per lb. What was the 


just share of the amount for each ? 


133. A and B hired a coach in Boston to go 50 miles, 7 
for $25. with liberty to take in two more when they — 


pleased. After riding 15 miles they took in C, who wished 


to go the remainder of the journey out, and return with — 


them to Boston. On their return, at the distance of 25 _ 
miles from Boston, they admitted D for the remainder ; 
of the journey. You are required to settle the coach hire — 


equitably between them. 
134. Suppose a rope 7.1365 rods long, to have one end 


attached to a horse’s head, and the other end fastened to 


a stake, in the centre of a field; how much land will the 
horse be allowed to graze upon ? 

135. A cistern is to be constructed, in the form of a 
cylinder, to hold 850 gallons. ‘If the diameter of the end 
be made 6 feet, what must be the length of the side? 


136. Suppose the propelling wheels of a locomotive — 


engine to be 3 feet 4 inches in diameter, and to make 396 


revolutions in a minute; what distance will the engine 


, 


move forward in one hour? 

137. If 12 oxen eat up 33 acres of grass in 4 weeks, 
and 21 oxen eat up 10 acres in 9 weeks, how many oxen 
will eat up 24 acres in 18 weeks; the grass being at first 
equal on every acre, and growing uniformly ? 


THE END. 


287 


' INDEX. 
Definitions of Quantity, Numbers, and Arithmetic -...-......... Ke 
Notation and Numeration «....-.-...seeceessecceccecveneees Poet. 6 
MU RIVISETOSTE. Lives ccm ad staat cles sae Reb Ub cM she cedadesccesete tues sos eudmar 9 
Subtraction «--«.ccssesccsees dette eeeeeneeeeeeense estes estes ee cces eres 10 
Multi plication ee lee otlas PN cae ca CME os ob Coiele ole cisaicine elthe c s ot slonientl ¢ 11 
DIVISION << oho. coco e ee cet ce sce ccc cece cccvenccceserstssesiowcusamins 14 
Properties of Numbers -..------2--++eeseseeteceeseeseecesseseeveceee 18 
BPPEIOTTIA NA Seite eocinrs duc sociedsc sees as oagiote ovgc en Ge Sese EO IMy aedele« 20 
Compound INT DEPS sual de Sain vc sass « tMiee o's « Cagataae ns a0 telae'elsis 95 
PPPCEHOTIB ee os oo oe elec elec cet ccencemestecctborservescsheasoveveadsers 36 
PPGEITAIO FUACTIONR «3 © caeseiss SSbs sec dssistewccsccet neccarts doen mnempy as 5l 
AT PATEe LGC RITIRIG: 6 toe 0 5 cc ceils anes vec’ ct sve cccpessec dedecncechamn’ 67 
PL CENSTIN OF NATITINO US cats fens ca larcea, nsec cence nb ewhvkcenesseapes ears 74 
Percentage .-....ecercerscccersececeeeeceseseneesesesesseercseenstens 86 
Se IMMOAIUTEG ds <ciececar ect sates cs cs esaet sac feccscocsccsesseateet castes 90 
PE des enon stele ns cede tes wes ercads anatase cen Sevedccedgecereeee ws e. OJ 
: _{nsurance -..-. eee ar yee Pep ae Ulites ec Veavaeugece gunn tun cols apa la eee 92 
Driter ie) cae - Cc laaeler da Satoh ne ch alee dbuwuadicndegs Gauss cheo dan nmams 93 
Partial Payments -..---++sesscsseecsesereseeeressseneseeeeerens Oeie ds 102 
Compound Interest ...--+-+.sseeseeeeeeeeeee pete eeeeeeneeeeeeeeeees 107 
BY Wieser a a A ep epee comes. ONE Rpts oc anne ares Sree Li 110 
Banking ....---0s0seseesescecesseeeecesereneeessscsesccsseessesesseres LIQ 
Equation of Payments --..+.+-.++seeeeeeeeeeeeeeeeeanec crete eenene es 114 
Pipceeanniche | ein ts ac ctdaeevieae so-so sicabloiss one cig cose ene stoves csunmige 116 
Partnership ---------+++++: OOF, os Stet sss score SHO RS Ue ete 117 
— Bankruptcy ----.-ceeceseee see ceceeeeneene eee ene steers snc see ecco ennes 121 
Assessmeht of TaxeS -----0csccccccccccsvsccscccsssssccscccvscesees 122 
General Average ---++-.esseereseereeeeeeecceeeeesenseenerecs essere. 124 
Custom-House Business....--.+...ssseccsesccecccecserevsescvecsers 127 
ENS 2 icon Fs chet E sacs NEE a's bay nce ae ciebein niin wales apes heigiepehin eles 137 
Proportion ....-2+-.seeeeeeeeeececeeceneeeesenseneeceen seer seeenenees 132 
Compound Proportion de Ms Os cans tle seven cal Ohlsson 140 
Conjoined Proportion «--.2+-.-+++sssereveerseeeeeecsee evens eenenss 145 
MP RSSCWAGRIVIDU: a's Sede cs oe be vlan Cana deenc sin t.cdes DSO APE cr 148 
DEF OREE LOTR soe eae ss vided sie bo ase evecsWis cwekep cesQueaniras ae 152 
BOMATILION,. cucbies cola cae sncuneeberncevedusdaseuae ddeedvasnsseempena ders 157 
Extraction of the Square Root ...+.+s+ssssesesereeeevenereerensees 16C 


- 
% 
Pi 
. : 


288 | INDEX. 
Extraction of the Cube Root -.........-.. UE Sena Er LEE = 174" 
Roots of All Powers s0ioescis < caed's yee gleieg wale = <°siaas niete thugs Ue 182 be 
Equidiffereht/Series »-.. 20.2 das... quacetan ne Sor eee Se er 184 : 
Continual Proportionals osha albve(e, »'S.Siukigie st oiats Taide gL ARMIN a MUAY ore Secs 188 © 
Compound Interest by Series: iiss de <0. tab akn ey ee ghee ces 194 : 
Compound: Discount — .....---seeeeeeees seveeceeeeescarserceeecees 195 
GAPSIU UR TONES oa ha = Lildig baie aethln' G's £3 & ssp to ge RO MML Claas » «5 aba ale a's pk irae bate al 196 
» Alligation” Medial =... 1... tyecssecend bases ge vinniagesonstivyseedecs 201 
Alligation A Ernatees 5 ses. Kiger s vet de B Oe o) se cee pene ane 203 
POMETIUITALONS Secon sos! ch Tiga ost eae Biss s ERM eee Set em acre tee 210 | 
Mombinatigns S-aweeke Meee es Meads «02 vee sv hanes tal ee ee ee 213 
BEX CHAN Ge Geen eke noo Age MR dy ad ow onity wel ADA coe a J16 
Great Britain .«.......:. eV aad Wate. eee see oe 939 
France te beeeeeceeereeees Boe PRONTO wudies > «scales sages 233 j 
Hamburgh eR as ce CY DE” Malta” “mrs sehen sence 992 4 
Amsterdam & Antwerp 223) Smyrna .--.---.+..s++-.-+s 234 
Portugal ......-.-...-0- Od. Caleitta . 2 dikes ste esa 935 
Spain i eee sor noes «aa 995 Bombay -+--++-.++-+se0e 926 7 
Swieeen es cc shus Ue cerned 226 Madras ..... abo Hee cee nc ORG 
FSIS ante lus, ween. O97... CantOtin stan: ted abies tous 237 : 
PUSS + - eins Monee coese as 228 Japan --...ceeeevenseeeees 237 
ED ETI VIALE -aisie.> » is Signe 40° v0 228 Sumatra vs dee ceie ssc mane 938 
Naples phew eee dc aite iii OSE. A cheetiiet.»ccacnusecteeuns 938 ‘ 
Sic ily Bede cai necdes aD SEM) oa Wey ies socials arcu dite se ree ee 239 . 
Leghorn .+s+.eeeeeeeeees LATTER sicasiteas vesoeecines 239 J 
GenOasss hotel eee Colombo0isscesate ose 939 P 
Wietiee siecese nawia sete IM AULEEILIS Sidcs chai ae ce 240 4 
Arbitration of Exchange «<i. cme... scaeeseesctenceescclanitdies ong 240 a 
Compound Arbitration -....-..-sscseecceseercsecceeveceeseneeree se 234 
“Foreign Coins seecseesceesecesccrccccersvesavccecencsascenerseereces DA5 | 
Foreign Weights and Measures. ..........sesessserceeeucetaverers 950 
AVE CYISUITSIIONL >. 9 oc ss-us ssitblers ce tinsic sb anieeth cc cena Pee GH. cli ecalciels 957 
MPAUSH OF CasKS cs Mee oss 0h ss. oa Mlle c5in, Pes cate os way meveaaie 963 
Tonnage of Vessels-...+....-+scsieescossetteescccerscsccsecegnmsores 964 — 
Mechanical Powers «5 .ies0c0seccccsite ssi ccc cdus sue Sine ty peo 266 © 
Miscellaneous Questions, sac» Mable. os. citenqenwanens <6 Veiga 975 


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